Practice Problems: Electric Potential Due to Point Charges Solutions
For all the problems below assume that V = 0 at infinity.

1. (easy) Determine the electric potential at 0.001 m from a charge of 2pC.
V = kq/r
V = 9x10⁹(2x10^-12)/(0.001) = 18 volts

2. (easy) Refer to the scenario in question #1.
a. If a second charge (-2pC) was the same distance from the point of interest as the first charge, find the total electric potential at that point.
The total potential is the algebraic sum of the potential caused by each charge:
V = Σkq/r
Since the distances are the same for each charge, but the sign is opposite, the total potential is zero in this case.
b. If the second charge was closer to the point of interest would the total potential be positive of negative?
Since the negative charge would be closer than the positive charge, the total potential would be negative.

3. (easy) Is the magnitude of the electric potential caused by point charges an absolute or a relative value. Explain your answer.
Electric potential is based on electric potential energy. The magnitude of the electric potential energy is relative to an established frame of reference. A convenient F.O.R. is usually utilized wherein a system of two charges separated by infinite distance has a potential energy of zero. Mathematically it can be shown that the electric potential infinitely far from a charge is zero when using the same F.O.R..

4. (moderate) Two charges are located on corners of a rectangle with a height of 0.05 m and a width of 0.15 m. The first charge (q₁= -5x10^-6 C) is located at the upper left hand corner, while the second charge (q₂ = +2.0 x10^-6 C) is at the lower right hand corner. Determine the electric potential at the upper right hand corner of the rectangle.
V = Σ(kq/r)
V = k[(-5x10^-6/0.15) + (2x10^-6/0.05)]
V = 60000 volts

5. (moderate) What is the potential difference for a point at the right hand corner (call it point A) of the rectangle in question #2 relative to the lower left hand corner (call it point B)?
V_A= 60000 volts
V_B = k[(-5x10^-6/0.05) + (2x10^-6/0.15)] = -780000 volts
ΔV = V_A – V_B = 60000 - (-780000) = 840000 volts

6. (moderate) Two charged particles are placed on the x axis of a coordinate system. The first (q₁ = 2x10^-6 C) is at the origin. The second (q₂ = -5x10^-6 C) is at x = 1.0 m. Determine a point in between these two charges where the electric potential is zero.
Let x be the distance from q₁ to the point of zero potential:
V = Σ(kq/r) = 0
0 = k[(2x10^-6/x) – (5x10^-6/(1 – x))]
18000/x = 45000/(1 – x)
x = 0.29 m

7. (moderate) Two charged particles are held in place on the x-axis of a coordinate system. Charge q₁ (5 C) is at the origin. Charge q₂ (3 C) is at x = 1 m. A relatively small positive test charge (q = 0.01 C, m = 0.001 kg) is released from rest at x = 0.5 m. Will the test charge move to the right or the left? Additionally, use the concepts of electric potential and electric potential energy to determine the speed of the test particle after it moves 0.1 m.
The test particle is repelled from both charges. It begins to move toward the 3C charge because the net force is in that direction.
To find the speed at x = 0.6 m, which is 0.1 m to the right of the initial position, we need to determine the change in K. The change in K is the opposite of the change in U. The change in U is related to the charge and the voltage.
V = k(q₁/r₁ + q₂/r₂)
ΔV = k[(5/0.6 + 3/0.4) - (5/0.5 + 3/0.5)]
ΔV = k(5.8) = -1.5x10⁹ volts
ΔU = qΔV
ΔU = (0.01)(-1.5x10⁹) = -1.5x10⁷ J
ΔK = -ΔU = 1.5x10⁷ J
ΔK = K₂ – K₁ = K₂ – 0
1.5x10⁷ = K₂ = ½mv²
1.5x10⁷ = ½(0.001)v²
v = 173205 m/s