Practice Problems: Equipotential Surface Solutions

1. (easy) A charged particle (q = 1.4 mC) moves 0.4 m along an equipotential surface of 10 volts. How much work is done by the field during this motion?
W = -ΔU
W = -qΔV
Since ΔV = 0 for movement along an equipotential surface, W = 0

2. (moderate) Two 1.0 C charges are at rest in a coordinate system. The first is negative and the second is positive. Their respective positions are (1.0 m, 1.0 m) and (1.0 m, 2.0 m). Determine the shape of an equipotential surface of which the points (1.0 m, 1.5 m) and (1.5 m, 1.5 m) are a part. Also, determine the magnitude of the potential on this surface.
Both of the positions are equidistant for both charges. Since the charges are equal and opposite, the potential at both points is zero. Any point on the equipotential surface of which these two are a part must have zero potential. The surface is thus a plane that bisects the system so that the addition of kq/r for each charge adds to zero.

3. (moderate) A positive particle (q = 1.0 C) is moving in a uniform E-field (E = 100 v/m) such that it is speeding up. The particle started from rest on an equipotential plane of V = 50 volts. After t = 0.0002 seconds the particle is on an equipotential plane of V = 10 volts. Determine the distance (d) the particle moved.
W = ΔK = -ΔU = -qΔV = -(1.0 C)(10 v – 50 v) = 40J
W = qEd
40 = qEd
40 = (1.0)(100)d
d = 0.4 m

4. (moderate) Answer the questions below based on your interpretation of the equipotential map shown here.


a. Which position, A or C, has a greater E-field? Explain.
The density of equipotentials (each 200 volts different from the others) is higher at position C than at position A. Since the potential gradient is directly related to the E-field, position C is in the region of higher E-field than position A.
b. Show the direction of the E-field at all four positions. Explain the reasoning for your answers.
The E-field is perpendicular to the equipotential line at each point on the various equipotentials. The direction of the field is toward the lower magnitude equipotentials. The E-field is always in the direction of lower potential.
c. If a proton was released from rest at position B, would it move toward the equipotential line of position A or position C? Explain.
Positive charges are pushed by the field in the direction of the field. Thus, since the field at position B is to the right, the charge will feel a force to the right. This is toward the equipotential line on which we see position C.
d. Repeat the previous question except assume the proton is now an electron. Would the electron gain or lose potential energy? Would the electron gain or lose electric potential?
The negative electron will be forced in the opposite direction of the field. It will move toward the equipotential line on which we find position A. The electron will lose potential energy and gain electric potential.
e. If a charged particle (q = 2 C) was moved by an external agent from position D to position B, calculate the work done by the agent and the work done by the field. Assume that the particle starts at rest and ends at rest.
W_agent = ΔU = qΔV
W_agent = 2(800 – 400) = 800 J
W_field = -W_agent = -800 J