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Voltage adds like numbers

Science · Electricity & Magnetism · ages 18-19

Name ______________________ Date ____________

A positive charge is released from rest in a fixed, nonzero electric field. If only the electric force acts, it initially accelerates toward lower potential.
Circle one: True False
What is electric potential?
- Potential energy per unit charge
- Electric force per unit test charge
- Electric charge passing per unit time
With zero potential at infinity, what is the potential a distance r > 0 from a point charge q?
- k times q over r squared
- k times q over r
- k times q divided by r cubed
A thin ring of radius 0.25 m has total charge +3 nC. Find the potential at its center, with zero at infinity. Use k = 9.0 × 10^9 N m²/C².
- 36 V
- 432 V
- 108 V
Charges +2 nC, -3 nC, +5 nC and -1 nC are fixed at a square's corners. Each is 0.20 m from its center. With V = 0 at infinity and k = 9.0 × 10^9 N m²/C², what is V at the center?
- 135 V
- 495 V
- -135 V
For a fixed electrostatic charge distribution, where may you choose zero potential?
- Only at a positive terminal, never at ground
- At a reference point where potential is finite and defined
- Only at infinity, even for an infinite line of charge
Two students use different zero references for the same fixed, finite charge distribution. One uses infinity; the other uses ground as a reference without changing any connections. Why do they predict the same motion for a charge with identical initial conditions?
- Both references differ by a constant, so the local electric field stays unchanged
- Ground zero changes the local field, so the charge must also change to keep its force and motion unchanged
- Infinity zero is physically required, so ground zero needs a compensating change in the charge's initial velocity
In a region, V(x) = 24 V - (6 V/m)x, with x in metres. What is the x component Ex of the electric field?
- 0 V/m
- 6 V/m
- -6 V/m

Answer key

For grown-ups. Fold this page away before handing over the rest.

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True · The electric force on a positive charge points along E, toward decreasing potential. This states the initial acceleration direction, not a rule for its whole later path.
Potential energy per unit charge · Electric potential is V = U/q. Its unit is J/C, called the volt. Force per charge defines electric field, and charge per time defines current.
k times q over r · For this reference, V = kq/r. Keep the sign of q. The point-charge electric field magnitude uses r squared instead.
108 V · At the center, every ring element is 0.25 m away. Thus V = k∫dq/R = kQ/R = (9.0 × 10^9)(3 × 10^-9)/0.25 = 108 V.
135 V · All four distances are equal. The signed total charge is +3 nC, so V = (9.0 × 10^9)(3 × 10^-9)/0.20 = 135 V.
At a reference point where potential is finite and defined · Choose a reference where potential is finite and defined. Subtracting its value adds a constant everywhere and preserves differences. A point-charge location is excluded because V = kq/r is undefined at r = 0.
Both references differ by a constant, so the local electric field stays unchanged · Adding a constant changes neither potential differences nor spatial derivatives. The local field E and force qE are unchanged, so identical initial conditions give the same motion.
6 V/m · The potential slope is -6 V/m. Since Ex = -∂V/∂x, the field component is +6 V/m, pointing along positive x.