To calculate the force exerted by some electric charges,
*q*_{1}, *q*_{2}, *q*_{3}, ... (**the
source charges**) on another charge *Q* (**the test charge**) we can
use the **principle of superposition**. This principle states that the
interaction between any two charges is completely unaffected by the presence of
other charges. The force exerted on *Q* by *q*_{1},
*q*_{2}, and *q*_{3} (see Figure 2.1) is therefore
equal to the vector sum of the force
_{}
exerted by *q*_{1} on *Q*, the force
_{}
exerted by *q*_{2} on *Q*, and the force
_{}
exerted by *q*_{3} on *Q*.

The force exerted by a charged particle on another charged particle depends
on their separation distance, on their velocities and on their accelerations.
In this Chapter we will consider the special case in which the source charges
are stationary.

The**electric field** produced by stationary source
charges is called and **electrostatic field**. The electric field at a
particular point is a vector whose magnitude is proportional to the total force
acting on a test charge located at that point, and whose direction is equal to
the direction of the force acting on a positive test charge. The electric field
_{},
generated by a collection of source charges, is defined as

The

where
_{}
is the total electric force exerted by the source charges on the test charge
*Q*. It is assumed that the test charge *Q* is small and therefore
does not change the distribution of the source charges. The total force exerted
by the source charges on the test charge is equal to

The electric field generated by the source charges is thus equal
to

In most applications the source charges are not discrete, but are
distributed continuously over some region. The following three different
distributions will be used in this course:

1.**line charge** λ:
the charge per unit length.

2.**surface charge** σ: the charge
per unit area.

3.**volume charge** ρ: the charge per unit
volume.

To calculate the electric field at a point_{}
generated by these charge distributions we have to replace the summation over
the discrete charges with an integration over the continuous charge
distribution:

1. for a line charge:_{}

2. for a surface charge:_{}

3. for a volume charge:_{}

Here_{}
is the unit vector from a segment of the charge distribution to the point
_{}
at which we are evaluating the electric field, and *r* is the distance
between this segment and point
_{}.

**Example:
Problem 2.2**

a) Find the electric field (magnitude and direction) a distance*z* above the midpoint between two equal charges *q* a
distance *d* apart. Check that your result is consistent with what you
would expect when *z* » *d*.

b) Repeat part a), only this time make he right-hand charge*-q* instead of +*q*.

1.

2.

3.

To calculate the electric field at a point

1. for a line charge:

2. for a surface charge:

3. for a volume charge:

Here

a) Find the electric field (magnitude and direction) a distance

b) Repeat part a), only this time make he right-hand charge

a) Figure 2.2a shows that the *x* components of the electric fields
generated by the two point charges cancel. The total electric field at *P
*is equal to the sum of the *z* components of the electric fields
generated by the two point charges:

When *z* » *d* this equation becomes approximately equal
to

which is the Coulomb field generated by a point charge with charge
2*q*.

b) For the electric fields generated by the point charges of the charge distribution shown in Figure 2.2b the*z* components cancel.
The net electric field is therefore equal to

b) For the electric fields generated by the point charges of the charge distribution shown in Figure 2.2b the

Find the electric field a distance

Each segment of the loop is located at the same distance from *P* (see
Figure 2.3). The magnitude of the electric field at *P* due to a segment
of the ring of length *dl* is equal to

When we integrate over the whole ring, the horizontal components of the
electric field cancel. We therefore only need to consider the vertical
component of the electric field generated by each segment:

The total electric field generated by the ring can be obtained by
integrating *dE*_{z} over the whole ring:

Find the electric field a distance

Consider a slice of the shell centered on the *z* axis (see Figure
2.4). The polar angle of this slice is *θ* and its width is
*dθ*. The area *dA* of this ring is

The total charge on this ring is equal to

where *q* is the total charge on the shell. The electric field
produced by this ring at *P* can be calculated using the solution of
Problem 2.5:

The total field at *P* can be found by integrating *dE* with
respect to *θ*:

This integral can be solved using the following relation:

Substituting this expression into the integral we obtain:

Outside the shell, *z* > *r* and consequently the electric
field is equal to

Inside the shell, *z* < *r* and consequently the electric
field is equal to

Thus the electric field of a charged shell is zero inside the shell. The
electric field outside the shell is equal to the electric field of a point
charge located at the center of the shell.

The electric field can be graphically represented using field lines. The
direction of the field lines indicates the direction in which a positive test
charge moves when placed in this field. The density of field lines per unit
area is proportional to the strength of the electric field. Field lines
originate on positive charges and terminate on negative charges. Field lines
can never cross since if this would occur, the direction of the electric field
at that particular point would be undefined. Examples of field lines produced
by positive point charges are shown in Figure 2.5.

The flux of electric field lines through any surface is proportional to the
number of field lines passing through that surface. Consider for example a
point charge *q* located at the origin. The electric flux
_{}
through a sphere of radius *r*, centered on the origin, is equal
to

Since the number of field lines generated by the charge *q* depends
only on the magnitude of the charge, any arbitrarily shaped surface that
encloses *q* will intercept the same number of field lines. Therefore the
electric flux through any surface that encloses the charge *q* is equal to
_{}.
Using the principle of superposition we can extend our conclusion easily to
systems containing more than one point charge:

We thus conclude that for an arbitrary surface and arbitrary charge
distribution

where *Q*_{enclosed} is the total charge enclosed by the
surface. This is called **Gauss's law**. Since this equation involves an
integral it is also called **Gauss's law in integral form**.

Using the divergence theorem the electric flux_{}
can be rewritten as

Using the divergence theorem the electric flux

We can also rewrite the enclosed charge *Q*_{encl} in terms of
the charge density *ρ*:

Gauss's law can thus be rewritten as

Since we have not made any assumptions about the integration volume this
equation must hold for any volume. This requires that the integrands are
equal:

This equation is called **Gauss's law in differential
form**.

Gauss's law in differential form can also be obtained directly from Coulomb's law for a charge distribution_{}:

Gauss's law in differential form can also be obtained directly from Coulomb's law for a charge distribution

where
_{}.
The divergence of
_{}
is equal to

which is Gauss's law in differential form. Gauss's law in integral form
can be obtained by integrating
_{}
over the volume *V*:

If the electric field in some region is given (in spherical coordinates) by the expression

where *A* and *B* are constants, what is the charge density
*ρ*?

The charge density*ρ* can be obtained from the
given electric field, using Gauss's law in differential form:

The charge density

Consider a charge distribution *ρ*(*r*). The electric
field at a point *P* generated by this charge distribution is equal
to

where
_{}.
The curl of
_{}
is equal to

However,
_{}
for every vector
_{}
and we thus conclude that

Although Gauss's law is always true it is only a useful tool to calculate
the electric field if the charge distribution is symmetric:

1. If the charge distribution has**spherical symmetry**, then Gauss's law can be used
with concentric spheres as Gaussian surfaces.

2. If the charge distribution has**cylindrical symmetry**, then Gauss's law can be used with
coaxial cylinders as Gaussian surfaces.

3. If the charge distribution has**plane symmetr**y, then Gauss's law can be used with pill boxes as Gaussian
surfaces.

**Example: Problem 2.12**

Use Gauss's law to find the electric field inside a uniformly charged sphere (charge density*ρ)
*of radius *R*.

The charge distribution has spherical symmetry and consequently the Gaussian surface used to obtain the electric field will be a concentric sphere of radius*r*. The electric flux through this surface
is equal to

1. If the charge distribution has

2. If the charge distribution has

3. If the charge distribution has

Use Gauss's law to find the electric field inside a uniformly charged sphere (charge density

The charge distribution has spherical symmetry and consequently the Gaussian surface used to obtain the electric field will be a concentric sphere of radius

The charge enclosed by this Gaussian surface is equal to

Applying Gauss's law we obtain for the electric field:

Find the electric field inside a sphere which carries a charge density proportional to the distance from the origin:

The charge distribution has spherical symmetry and we will therefore use a concentric sphere of radius

The charge enclosed by the Gaussian surface can be obtained by integrating
the charge distribution between *r'* = 0 and *r'* =
*r*:

Applying Gauss's law we obtain:

or

A long coaxial cable carries a uniform (positive) volume charge density

The charge distribution has cylindrical symmetry and to apply Gauss's law we will use a cylindrical Gaussian surface. Consider a cylinder of radius

The enclosed charge must be calculated separately for each of the three
regions:

1.*r* <
*a*: _{}

2.*a*
< *r* <
*b*: _{}

3.*b*
<
*r*: _{}

Applying Gauss's law we find

1.

2.

3.

Applying Gauss's law we find

Substituting the calculated *Q*_{encl} for the three regions
we obtain

1.*r* <
*a*: _{}.

2.*a*
< *r* <
*b*: _{}

3.*b*
<
*r* _{}

**Example:
Problem 2.18**

Two spheres, each of radius*R* and carrying uniform
charge densities of +*ρ* and -*ρ*, respectively, are placed
so that they partially overlap (see Figure 2.6). Call the vector from the
negative center to the positive center
_{}.
Show that the field in the region of overlap is constant and find its
value.

To calculate the total field generated by this charge distribution we use the principle of superposition. The electric field generated by each sphere can be obtained using Gauss' law (see Problem 2.12). Consider an arbitrary point in the overlap region of the two spheres (see Figure 2.7). The distance between this point and the center of the negatively charged sphere is*r*_{-}. The distance between this point and the center
of the positively charged sphere is *r*_{+}. Figure 2.7 shows that
the vector sum of
_{}
and
_{}
is equal to
_{}.
Therefore,

1.

2.

3.

Two spheres, each of radius

To calculate the total field generated by this charge distribution we use the principle of superposition. The electric field generated by each sphere can be obtained using Gauss' law (see Problem 2.12). Consider an arbitrary point in the overlap region of the two spheres (see Figure 2.7). The distance between this point and the center of the negatively charged sphere is

The total electric field at this point in the overlap region is the vector
sum of the field due to the positively charged sphere and the field due to the
negatively charged sphere:

The minus sign in front of
_{}
shows that the electric field generated by the negatively charged sphere is
directed opposite to
_{}.
Using the relation between
_{}
and
_{}
obtained from Figure 2.7 we can rewrite
_{}
as

which shows that the field in the overlap region is homogeneous and
pointing in a direction opposite to
_{}.

The requirement that the curl of the electric field is equal to zero
limits the number of vector functions that can describe the electric field. In
addition, a theorem discussed in Chapter 1 states that any vector function whose
curl is equal to zero is the gradient of a scalar function. The scalar function
whose gradient is the electric field is called the **electric potential
***V* and it is defined as

Taking the line integral of
_{}
between point *a* and point *b* we obtain

Taking *a* to be the reference point and defining the potential to be
zero there, we obtain for *V*(*b*)

The choice of the reference point *a* of the potential is arbitrary.
Changing the reference point of the potential amounts to adding a constant to
the potential:

where *K* is a constant, independent of *b*, and equal
to

However, since the gradient of a constant is equal to zero

Thus, the electric field generated by *V'* is equal to the electric
field generated by *V*. The physical behavior of a system will depend only
on the difference in electric potential and is therefore independent of the
choice of the reference point. The most common choice of the reference point in
electrostatic problems is infinity and the corresponding value of the potential
is usually taken to be equal to zero:

The unit of the electrical potential is the Volt (V, 1V = 1
Nm/C).

**Example: Problem 2.20**

One of these is an impossible electrostatic field. Which one?

a)_{}

b)_{}

Here,*k* is a constant with the appropriate units. For the *possible* one,
find the potential, using the origin as your reference point. Check your answer
by computing
_{}.

a) The curl of this vector function is equal to

One of these is an impossible electrostatic field. Which one?

a)

b)

Here,

a) The curl of this vector function is equal to

Since the curl of this vector function is not equal to zero, this vector
function can not describe an electric field.

b) The curl of this vector function is equal to

b) The curl of this vector function is equal to

Since the curl of this vector function is equal to zero it can describe an
electric field. To calculate the electric potential *V* at an arbitrary
point (*x*, *y*, *z*), using (0, 0, 0) as a reference point, we
have to evaluate the line integral of
_{}
between (0, 0, 0) and (*x*, *y*, *z*). Since the line integral
of
_{}
is path independent we are free to choose the most convenient integration path.
I will use the following integration path:

The first segment of the integration path is along the *x*
axis:

and

since *y* = 0 along this path. Consequently, the line integral of
_{}
along this segment of the integration path is equal to zero. The second segment
of the path is parallel to the *y* axis:

and

since *z* = 0 along this path. The line integral of
_{}
along this segment of the integration path is equal to

The third segment of the integration path is parallel to the *z*
axis:

and

The line integral of
_{}
along this segment of the integration path is equal to

The electric potential at (*x*, *y*, *z*) is thus equal
to

The answer can be verified by calculating the gradient of
*V*:

which is the opposite of the original electric field
_{}.

The advantage of using the electric potential*V* instead of the electric field
is that *V* is a scalar function. The total electric potential generated
by a charge distribution can be found using the superposition principle. This
property follows immediately from the definition of *V* and the fact that
the electric field satisfies the principle of superposition. Since

The advantage of using the electric potential

it follows that

This equation shows that the total potential at any point is the algebraic
sum of the potentials at that point due to all the source charges separately.
This ordinary sum of scalars is in general easier to evaluate then a vector
sum.

**Example: Problem 2.46**

Suppose the electric potential is given by the expression

Suppose the electric potential is given by the expression

for all *r* (*A* and *λ* are constants). Find the
electric field
_{},
the charge density
_{},
and the total charge *Q*.

The electric field_{}
can be immediately obtained from the electric potential:

The electric field

The charge density
_{}
can be found using the electric field
_{}
and the following relation:

This expression shows that

Substituting the expression for the electric field
_{}
we obtain for the charge density
_{}:

The total charge *Q* can be found by volume integration of
_{}:

The integral can be solved easily:

The total charge is thus equal to

The charge distribution

This equation can be rewritten as

and is known as **Poisson's equation**. In the regions where
_{}
this equation reduces to **Laplace's equation**:

The electric potential generated by a discrete charge distribution can be
obtained using the principle of superposition:

where
_{}
is the electric potential generated by the point charge
_{}.
A point charge
_{}
located at the origin will generate an electric potential
_{}
equal to

In general, point charge
_{}
will be located at position
_{}
and the electric potential generated by this point charge at position
_{}
is equal to

The total electric potential generated by the whole set of point charges is
equal to

To calculate the electric potential generated by a continuous charge
distribution we have to replace the summation over point charges with an
integration over the continuous charge distribution. For the three charge
distributions we will be using in this course we obtain:

1. line charge*λ *:
_{}

2. surface charge*σ *:
_{}

3. volume charge*ρ *:
_{}

**Example:
Problem 2.25**

Using the general expression for*V* in terms of
*ρ* find the potential at a distance *z* above the center of the
charge distributions of Figure 2.8. In each case, compute
_{}.
Suppose that we changed the right-hand charge in Figure 2.8a to -*q*. What
is then the potential at *P*? What field does this suggest? Compare your
answer to Problem 2.2b, and explain carefully any discrepancy.

1. line charge

2. surface charge

3. volume charge

Using the general expression for

a) The electric potential at *P* generated by the two point charges is
equal to

The electric field generated by the two point charges can be obtained by
taking the gradient of the electric potential:

If we change the right-hand charge to -*q* then the total potential at
*P* is equal to zero. However, this does not imply that the electric field
at *P* is equal to zero. In our calculation we have assumed right from the
start that *x* = 0 and *y* = 0. Obviously, the potential at *P*
will therefore not show an *x* and *y* dependence. This however not
necessarily indicates that the components of the electric field along the
*x* and *y* direction are zero. This can be demonstrated by
calculating the general expression for the electric potential of this charge
distribution at an arbitrary point (*x*,*y*,*z*):

The various components of the electric field can be obtained by taking the
gradient of this expression:

The components of the electric field at *P* = (0, 0, *z*) can now
be calculated easily:

b) Consider a small segment of the rod, centered at position *x* and
with length *dx*. The charge on this segment is equal to
_{}.
The potential generated by this segment at *P* is equal to

The total potential generated by the rod at *P* can be obtained by
integrating *dV* between *x* = - *L* and *x* =
*L*

The *z* component of the electric field at *P* can be obtained
from the potential *V* by calculating the *z* component of the
gradient of *V*. We obtain

c) Consider a ring of radius *r* and width *dr*. The charge on
this ring is equal to

The electric potential *dV* at *P* generated by this ring is
equal to

The total electric potential at *P* can be obtained by integrating
*dV* between *r* = 0 and *r* = *R*:

The *z* component of the electric field generated by this charge
distribution can be obtained by taking the gradient of *V*:

Find the electric field a distance

The total charge

The total electric potential *V* at *P* is equal to

The *z* component of the electric field at *P* can be obtained by
calculating the gradient of *V*:

This is the same answer we obtained in the beginning of this Chapter by
taking the vector sum of the segments of the ring.

We have seen so far that there are three fundamental quantities of electrostatics:

1. The**charge density** *ρ*

2. The**electric field**
_{}

3. The**electric potential** *V*

If one of these quantities is known, the others can be calculated:

We have seen so far that there are three fundamental quantities of electrostatics:

1. The

2. The

3. The

If one of these quantities is known, the others can be calculated:

In general the charge density *ρ* and the electric field
_{}
do not have to be continuous. Consider for example an infinitesimal thin charge
sheet with surface charge *σ*. The relation between the electric
field above and below the sheet can be obtained using Gauss's law. Consider a
rectangular box of height *ε* and area *A* (see Figure 2.9). The
electric flux through the surface of the box, in the limit *ε* →
0, is equal to

where
_{}
and
_{}
are the perpendicular components of the electric field above and below the
charge sheet. Using Gauss's law and the rectangular box shown in Figure 2.9 as
integration volume we obtain

This equation shows that the electric field perpendicular to the charge
sheet is discontinuous at the boundary. The difference between the
perpendicular component of the electric field above and below the charge sheet
is equal to

The tangential component of the electric field is always continuous at any
boundary. This can be demonstrated by calculating the line integral of
_{}
around a rectangular loop of length *L* and height *ε* (see
Figure 2.10). The line integral of
_{},
in the limit *ε* → 0, is equal to

Since the line integral of
_{}
around any closed loop is zero we conclude that

or

These boundary conditions for
_{}
can be combined into a single formula:

where
_{}
is a unit vector perpendicular to the surface and pointing towards the
*above* region.

The electric potential is continuous across any boundary. This is a direct results of the definition of*V* in terms of
the line integral of
_{}:

The electric potential is continuous across any boundary. This is a direct results of the definition of

If the path shrinks the line integral will approach zero, independent of
whether
_{}
is continuous or discontinuous. Thus

a) Check that the results of examples 4 and 5 of Griffiths are consistent with the boundary conditions for

b) Use Gauss's law to find the field inside and outside a long hollow cylindrical tube which carries a uniform surface charge

c) Check that the result of example 7 of Griffiths is consistent with the boundary conditions for

a)

Therefore,

which is in agreement with the boundary conditions for
_{}.

__Example
5 (Griffiths)__: The electric field generated by the two charge sheets is
directed perpendicular to the sheets and has a magnitude equal to

The change in the strength of the electric field at the left sheet is equal
to

The change in the strength of the electric field at the right sheet is
equal to

These relations show agreement with the boundary conditions for
_{}.

b) Consider a Gaussian surface of length*L* and radius *r*. As a result of the
symmetry of the system, the electric field will be directed radially. The
electric flux through this Gaussian surface is therefore equal to the electric
flux through its curved surface which is equal to

b) Consider a Gaussian surface of length

The charge enclosed by the Gaussian surface is equal to zero when *r*
< *R*. Therefore

when *r* < *R*. When *r* > *R* the charge
enclosed by the Gaussian surface is equal to

The electric field for *r* > *R*, obtained using Gauss' law,
is equal to

The magnitude of the electric field just outside the cylinder, directed
radially, is equal to

The magnitude of the electric field just inside the cylinder is equal
to

Therefore,

which is consistent with the boundary conditions for
*E*.

c)__Example 7 (Griffiths)__: the electric potential
just outside the charged spherical shell is equal to

c)

The electric potential just inside the charged spherical shell is equal
to

These two equations show that the electric potential is continuous at the
boundary.

Consider a point charge *q*_{1} located at the origin. A
point charge *q*_{2} is moved from infinity to a point a distance
*r*_{2} from the origin. We will assume that the point charge
*q*_{1} remains fixed at the origin when point charge
*q*_{2} is moved. The force exerted by *q*_{1} on
*q*_{2} is equal to

where
_{}
is the electric field generated by *q*_{1}. In order to move
charge *q*_{2} we will have to exert a force opposite to
_{}.
Therefore, the total work that must be done to move *q*_{2} from
infinity to *r*_{2} is equal to

where
_{}
is the electric potential generated by *q*_{1} at position
*r*_{2}. Using the equation of *V* for a point charge, the
work *W* can be rewritten as

This work *W* is the work necessary to assemble the system of two
point charges and is also called the __electrostatic potential
energy__ of the system. The energy of a system of more than two point
charges can be found in a similar manner using the superposition principle. For
example, for a system consisting of three point charges (see Figure 2.11) the
electrostatic potential energy is equal to

In this equation we have added the electrostatic energies of each pair of
point charges. The general expression of the electrostatic potential energy for
*n* point charges is

The lower limit of *j* (= *i* + 1) insures that each pair of
point charges is only counted once. The electrostatic potential energy can also
be written as

where *V*_{i} is the electrostatic potential at the location
of *q*_{i} due to all other point charges.

When the charge of the system is not distributed as point charges, but rather as a continuous charge distribution*ρ*, then the electrostatic potential energy of
the system must be rewritten as

When the charge of the system is not distributed as point charges, but rather as a continuous charge distribution

For continuous surface and line charges the electrostatic potential energy
is equal to

and

However, we have already seen in this Chapter that *ρ*, *V*,
and
_{}
carry the same equivalent information. The charge density *ρ*, for
example, is related to the electric field
_{}:

Using this relation we can rewrite the electrostatic potential energy
as

where we have used one of the product rules of vector derivatives and the
definition of
_{}
in terms of *V*. In deriving this expression we have not made any
assumptions about the volume considered. This expression is therefore valid for
any volume. If we consider all space, then the contribution of the surface
integral approaches zero since
_{}
will approach zero faster than 1/*r*^{2}. Thus the total
electrostatic potential energy of the system is equal to

A sphere of radius

The first method we will use to calculate the electrostatic potential energy of the charged sphere uses the volume integral of

The electric flux through the Gaussian surface is equal to

Applying Gauss's law we find for the electric field inside the sphere
(*r* < *R*):

The electric field outside the sphere (*r* > *R*) can also be
obtained using Gauss's law:

The total electrostatic energy can be obtained from the electric
field:

An alternative way calculate the electrostatic potential energy is to use the following relation:

The electrostatic potential *V* can be obtained immediately from the
electric field
_{}.
To evaluate the volume integral of
_{}
we only need to know the electrostatic potential *V* inside the charged
sphere:

The electrostatic potential energy of the system is thus equal to

which is equal to the energy calculated using method 1.

In a **metallic conductor** one or more electrons per atom are free to
move around through the material. Metallic conductors have the following
electrostatic properties:

1.**The electric field inside the conductor
is equal to zero.**

If there would be an electric field inside the conductor, the free charges would move and produce an electric field of their own opposite to the initial electric field. Free charges will continue to flow until the cancellation of the initial field is complete.

2.**The charge
density inside a conductor is equal to zero.**

This property is a direct result of property 1. If the electric field inside a conductor is equal to zero, then the electric flux through any arbitrary closed surface inside the conductor is equal to zero. This immediately implies that the charge density inside the conductor is equal to zero everywhere (Gauss's law).

3.**Any
net charge of a conductor resides on the surface.**

Since the charge density inside a conductor is equal to zero, any net charge can only reside on the surface.

4.**The electrostatic potential ***V* is constant
throughout the conductor.

Consider two arbitrary points*a* and
*b* inside a conductor (see Figure 2.12). The potential difference between
*a* and *b* is equal to

1.

If there would be an electric field inside the conductor, the free charges would move and produce an electric field of their own opposite to the initial electric field. Free charges will continue to flow until the cancellation of the initial field is complete.

2.

This property is a direct result of property 1. If the electric field inside a conductor is equal to zero, then the electric flux through any arbitrary closed surface inside the conductor is equal to zero. This immediately implies that the charge density inside the conductor is equal to zero everywhere (Gauss's law).

3.

Since the charge density inside a conductor is equal to zero, any net charge can only reside on the surface.

4.

Consider two arbitrary points

Since the electric field inside a conductor is equal to zero, the line
integral of
_{}
between *a* and *b* is equal to zero. Thus

or

5. **The electric field is perpendicular to the surface, just outside the
conductor.**

If there would be a tangential component of the electric field at the surface, then the surface charge would immediately flow around the surface until it cancels this tangential component.

**Example: A
spherical conducting shell**

a) Suppose we place a point charge*q* at
the center of a neutral spherical conducting shell (see Figure 2.13). It will
attract negative charge to the inner surface of the conductor. How much induced
charge will accumulate here?

b) Find*E* and *V* as function of
*r* in the three regions *r* < *a*, *a* < *r*
< *b*, and *r* > *b*.

If there would be a tangential component of the electric field at the surface, then the surface charge would immediately flow around the surface until it cancels this tangential component.

a) Suppose we place a point charge

b) Find

a) The electric field inside the conducting shell is equal to zero
(property 1 of conductors). Therefore, the electric flux through any concentric
spherical Gaussian surface of radius *r* (*a*<*r*<*b*)
is equal to zero. However, according to Gauss's law this implies that the
charge enclosed by this surface is equal to zero. This can only be achieved if
the charge accumulated on the inside of the conducting shell is equal to
-*q*. Since the conducting shell is neutral and any net charge must reside
on the surface, the charge on the outside of the conducting shell must be equal
to +*q*.

b) The electric field generated by this system can be calculated using Gauss's law. In the three different regions the electric field is equal to

_{} for
*b* <
*r*

_{} for
*a* < *r* <
*b*

_{} for
*r* < *a*

The electrostatic potential*V*(*r*) can
be obtained by calculating the line integral of
_{}
from infinity to a point a distance *r* from the origin. Taking the
reference point at infinity and setting the value of the electrostatic potential
to zero there we can calculate the electrostatic potential. The line integral
of
_{}
has to be evaluated for each of the three regions separately.

For*b*
< *r*:

b) The electric field generated by this system can be calculated using Gauss's law. In the three different regions the electric field is equal to

The electrostatic potential

For

For *a* < *r* < *b*:

For *r* < *a*:

In this example we have looked at a symmetric system but the general
conclusions are also valid for an arbitrarily shaped conductor. For example,
consider the conductor with a cavity shown in Figure 2.14. Consider also a
Gaussian surface that completely surrounds the cavity (see for example the
dashed line in Figure 2.14). Since the electric field inside the conductor is
equal to zero, the electric flux through the Gaussian surface is equal to zero.
Gauss's law immediately implies that the charge enclosed by the surface is equal
to zero. Therefore, if there is a charge *q* inside the cavity there will
be an induced charge equal to -*q* on the walls of the cavity. On the
other hand, if there is no charge inside the cavity then there will be no charge
on the walls of the cavity. In this case, the electric field inside the cavity
will be equal to zero. This can be demonstrated by assuming that the electric
field inside the cavity is not equal to zero. In this case, there must be at
least one field line inside the cavity. Since field lines originate on a
positive charge and terminate on a negative charge, and since there is no charge
inside the cavity, this field line must start and end on the cavity walls (see
for example Figure 2.15). Now consider a closed loop, which follows the field
line inside the cavity and has an arbitrary shape inside the conductor (see
Figure 2.15). The line integral of
_{}
inside the cavity is definitely not equal to zero since the magnitude of
_{}
is not equal to zero and since the path is defined such that
_{}
and
_{}
are parallel. Since the electric field inside the conductor is equal to zero,
the path integral of
_{}
inside the conductor is equal to zero. Therefore, the path integral of
_{}
along the path indicated in Figure 2.15 is not equal to zero if the magnitude of
_{}
is not equal to zero inside the cavity. However, the line integral of
_{}
along any closed path must be equal to zero and consequently the electric field
inside the cavity must be equal to zero.

A metal sphere of radius

a) Find the surface charge density

b) Find the potential at the center of the sphere, using infinity as reference.

c) Now the outer surface is touched to a grounding wire, which lowers its potential to zero (same as at infinity). How do your answers to a) and b) change?

a) Since the net charge of a conductor resides on its surface, the charge

As a result of the charge on the metal sphere there will be a charge equal
to -*q* induced on the inner surface of the metal shell. Its surface
charge density will therefore be equal to

Since the metal shell is neutral there will be a charge equal to +*q*
on the outside of the shell. The surface charge density on the outside of the
shell will therefore be equal to

b) The potential at the center of the metal sphere can be found by
calculating the line integral of
_{}
between infinity and the center. The electric field in the regions outside the
sphere and shell can be found using Gauss's law. The electric field inside the
shell and sphere is equal to zero. Therefore,

c) When the outside of the shell is grounded, the charge density on the
outside will become zero. The charge density on the inside of the shell and on
the metal sphere will remain the same. The electric potential at the center of
the system will also change as a result of grounding the outer shell. Since the
electric potential of the outer shell is zero, we do not need to consider the
line integral of
_{}
in the region outside the shell to determine the potential at the center of the
sphere. Thus

Consider a conductor with surface charge

This electric field will exert a force on the surface charge.

Consider a small, infinitely thin, patch of the surface with surface area*dA* (see
Figure 2.17). The electric field directly above and below the patch is equal to
the vector sum of the electric field generated by the patch, the electric field
generated by the rest of the conductor and the external electric field. The
electric field generated by the patch is equal to

Consider a small, infinitely thin, patch of the surface with surface area

The remaining field,
_{},
is continuous across the patch, and consequently the total electric field above
and below the patch is equal to

These two equations show that
_{}
is equal to

In this case the electric field below the surface is equal to zero and the
electric field above the surface is directly determined by the boundary
condition for the electric field at the surface. Thus

Since the patch cannot exert a force on itself, the electric force exerted
on it is entirely due to the electric field
_{}.
The charge on the patch is equal to
_{}.
Therefore, the force exerted on the patch is equal to

The force per unit area of the conductor is equal to

This equation can be rewritten in terms of the electric field just outside
the conductor as

This force is directed outwards. It is called the **radiation
pressure**.

Consider two conductors (see Figure 2.18), one with a charge equal to
+*Q* and one with a charge equal to -*Q*. The potential difference
between the two conductors is equal to

Since the electric field
_{}
is proportional to the charge *Q*, the potential difference ∆*V*
will also be proportional to *Q*. The constant of proportionality is
called the **capacitance ***C* of the system and is defined as

The capacitance *C* is determined by the size, the shape, and the
separation distance of the two conductors. The unit of capacitance is the
**farad (***F*). The capacitance of a system of conductors can in
general be calculated by carrying out the following steps:

1. Place a charge +*Q* on one of the conductors. Place a charge of -*Q* on the other
conductor (for a two conductor system).

2. Calculate the electric field in the region between the two conductors.

3. Use the electric field calculated in step 2 to calculate the potential difference between the two conductors.

4. Apply the result of part 3 to calculate the capacitance:

1. Place a charge +

2. Calculate the electric field in the region between the two conductors.

3. Use the electric field calculated in step 2 to calculate the potential difference between the two conductors.

4. Apply the result of part 3 to calculate the capacitance:

We will now discuss two examples in which we follow these steps to
calculate the capacitance.

**Example: Example 2.11
(Griffiths)**

Find the capacitance of two concentric shells, with radii*a* and *b*.

Place a charge*+Q* on the inner shell and a
charge -*Q* on the outer shell. The electric field between the shells can
be found using Gauss's law and is equal to

Find the capacitance of two concentric shells, with radii

Place a charge

The potential difference between the outer shell and the inner shell is
equal to

The capacitance of this system is equal to

A system does not have to have two conductors in order to have a
capacitance. Consider for example a single spherical shell of radius *R*.
The capacitance of this system of conductors can be calculated by following the
same steps as in Example 12. First of all, put a charge *Q* on the
conductor. Gauss's law can be used to calculate the electric field generated by
this system with the following result:

Taking infinity as the reference point we can calculate the electrostatic
potential on the surface of the shell:

Therefore, the capacitance of the shell is equal to

Let us now consider a parallel-plate capacitor. The work required to
charge up the parallel-plate capacitor can be calculated in various
ways:

__Method 1__: Since we are free to chose the reference
point and reference value of the potential we will chose it such that the
potential of the positively charges plate is
_{}
and the potential of the negatively charged plate is
_{}.
The energy of this charge distribution is then equal to

To increase the charge on the positively charged conductor by *dq* we
have to move this charge *dq* across this potential difference
_{}.
The work required to do this is equal to

Therefore, the total work required to charge up the capacitor from *q*
= 0 to *q* = *Q* is equal to

Suppose the plates of a parallel-plate capacitor move closer together by an infinitesimal distance

a) Use equation (2.45) of Griffiths to express the amount of work done by electrostatic forces, in terms of the field

b) Use equation (2.40) of Griffiths to express the energy lost by the field in this process.

a) We will assume that the parallel-plate capacitor is an ideal capacitor with a homogeneous electric field

The total force exerted on each plate is therefore equal to

As a result of this force, the plates of the parallel-plate capacitor move
closer together by an infinitesimal distance *ε*. The work done by
the electrostatic forces during this movement is equal to

b) The total energy stored in the electric field is equal to

In an ideal capacitor the electric field is constant between the plates and
consequently we can easily evaluate the volume integral of
_{}:

where *d* is the distance between the plates. If the distance between
the plates is reduced, then the energy stored in the field will also be reduced.
A reduction in *d* of *ε* will reduce the energy stored by an
amount ∆*W* equal to

which is equal to the work done by the electrostatic forces on the
capacitor plates (see part a).