Solutions Physics 122 Midterm Exam # 1


Problem 1

Figure Problem 1

Consider the Figure above which shows the electric field generated by two point charges. The following statements can be made in connection with the Figure:


Problem 2

Figure Problem 2


Problem 3

Consider an infinite rectangular box of height 2d as a Gaussian surface, and assume that the charge Q is located in the center of that box. Using Gauss' law we can conclude that the flux through this box is equal to

Due to the symmetry of the problem, half of the flux will go through the top infinite plane of the box, and half of it will go through the bottom infinite plane of the box. Thus the flux through each plane is equal to


Problem 4

Figure Problem 4

This problem can be solved using the principle of superposition. Consider first a solid sphere of radius R and charge Q. The electric field generated by this charge distribution at point P is equal to

Now consider a second sphere with a charge density, identical in magnitude but of opposite sign, and a diameter R, located a distance R/2 from the center of the solid sphere. The electric field produced by this sphere at P is equal to

The superposition of these two charge distributions produces the charge distribution shown in this problem. The electric field generated by this charge distribution is thus the superposition of the electric fields generated by each of these two charge distributions separately.


Problem 5

The electric fields inside the inner shell and outside the outer shells are equal to zero N/C. The electric field between the shells are that of a point charge Q (where Q is equal to the charge on the inner shell) located at the center of the shells:

The corresponding energy density u(r) is equal to

The total energy stored in the system can be obtained by integrating the energy density over the volume between the shells:


Problem 6

Figure Problem 6

The following conclusions can be drawn for electric field generated by the charge configuration shown in the Figure:


Problem 7

Figure Problem 7

Due to the symmetry of the problem we can immediately conclude that the net electric field on the axis of the ring is directed along this axis (see Figure). Consider a small segment of the ring with charge dQ. The electric field generate on the axis of the ring by this charge dQ is equal to

The component of the field directed along the axis of the ring is equal to

The total electric field due to the entire ring is thus equal to

The problem specifies the charge density of the ring. Using the specified density we conclude that


Problem 8

We can calculate the potential energy of the system most easily by using the following equation:

The potential at z1 due to the other two charges is equal to

The potential at z2 due to the other two charges is equal to

The potential at z3 due to the other two charges is equal to

The total potential energy of the system is thus equal to