Electric fields and potential: keep the charge sign

MCAT content library / Chemistry & physics / Electric fields and potential: keep the charge sign

The field can point right while a negative charge accelerates left. A voltage map and the sign of the charge make both force and energy predictions consistent.

About 20 minutes, including the worked examples and checks.

Useful foundations

On this page

Keep the source, the field, and the test charge separate
Use a vector sum for forces and fields
Potential is a scalar; potential energy includes the test charge
Worked application: one region, two signs of charge
Compare paths by endpoints, and motion by the forces
Check your reasoning
Continue learning
Sources

Keep the source, the field, and the test charge separate

An electric field describes how a specified arrangement of source charges would act on a small test charge. Replacing a positive test charge with a negative one changes the force on that charge without reversing the source field. Electric potential describes the same environment through an energy-per-charge account. AAMC category 4C includes both representations and their relationship to electrostatic forces.¹

Electric charge is conserved when an isolated system is considered as a whole. Charging by transfer redistributes charge; it does not create a net excess from nothing. A conductor permits charge to move relatively readily, whereas an insulator restricts that long-range movement. An isolated conductor can still hold a net charge. “Conductor” describes mobility, not an automatic absence of charge.²³

Use a vector sum for forces and fields

For two point charges in vacuum, the force magnitude is F = k|Qq|/r², with k approximately 9.0×10⁹N·m²/C². Like signs repel and unlike signs attract along the line joining them. Doubling separation reduces the magnitude to one quarter; it does not change attraction into repulsion.

At a position r from one point source Q, the field magnitude is E = k|Q|/r². It points away from a positive source and toward a negative source. For multiple sources, add their fields as vectors at the position of interest. Only then use F = qE, retaining the sign of the test charge. Field units N/C are equivalent to V/m.⁴

As an original numerical example, a +3.0 nC source produces E = (9.0×10⁹)(3.0×10⁻⁹)/(0.30)² =300 N/C at 0.30 m. A −2.0 nC test charge there experiences a force of magnitude 6.0×10⁻⁷N directed toward the source. The field points outward, while this negative charge’s force points inward.

Potential is a scalar; potential energy includes the test charge

Electric potential V is measured in volts, where 1 V =1 J/C. Between two points, ΔU = qΔV. The electrostatic force does work W_electric = −ΔU = −qΔV. The arbitrary location assigned zero potential does not affect either difference. A charge’s potential energy can decrease while the potential it moves toward increases, because q can be negative.⁵

For a point source with zero potential at infinity, V = kQ/r. Add signed potentials as scalars when there are several sources. Equal positive sources on opposite sides of a midpoint give cancelling fields there but positive summed potential. Field cancellation and potential cancellation are therefore different tests.⁶

Worked application: one region, two signs of charge

The figure, values, and tables are original illustrative models. Large parallel plates are 0.060 m apart, with the left plate at 120 V and the right plate at 0 V. Neglect edge effects and changes caused by the very small test charges. The field in the central region is uniform.

Original ideal parallel-plate model, away from edge effects. Dashed lines mark equal-potential planes. The lower charge arrows compare two separate test charges at the same position; their drawings are not additional source charges in the plate model.

Distance from left plate (m)	Potential (V)
0	120
0.015	90
0.030	60
0.045	30
0.060	0

Potential across the ideal plate gap

The potential falls 120 V over 0.060 m, so the field magnitude is 120/0.060 =2000 V/m and its direction is right. In a one-dimensional uniform field, E_x = −ΔV/Δx. The minus sign connects a positive rightward field to decreasing potential as x increases. More closely spaced equal-voltage contours would indicate a larger field magnitude in that direction.⁷

Move a +2.0 nC charge from the 90 V plane to the 30 V plane. Its potential change is−60 V, so ΔU = (2.0×10⁻⁹)(−60) =−120 nJ. The field does+120 nJ of work. If that is the only work, kinetic energy rises by 120 nJ. The force magnitude is qE =4.0 µN.

Move a −2.0 nC charge between the same endpoints. Now ΔU =+120 nJ and the field does−120 nJ of work. The force points left while the imposed displacement is right. A separate agent can make that transfer, but an initially stationary negative charge acted on only by this electric field would initially accelerate left, toward higher V and lower U.

Test charge	ΔU	Work by electric field	Force direction
+2.0 nC	−120 nJ	+120 nJ	Right
−2.0 nC	+120 nJ	−120 nJ	Left

Same90 V→30 V transfer, different test-charge sign

Compare paths by endpoints, and motion by the forces

In electrostatics with fixed sources, electric work between two points depends on their potential difference, not the shape or length of the route. A detour across this plate region therefore leaves the total electric work unchanged if the endpoints remain the same. Other forces can do different work along different routes, so equality of electric work does not automatically imply equality of every energy transfer.⁸

Movement along an equipotential produces no electric potential-energy change for a fixed q. That does not require the field itself to be zero: the displacement can be perpendicular to the field. Nor does a potential map alone specify a moving particle’s entire trajectory; initial velocity and any other forces still matter.

A reliable sequence is to identify the source field or potential difference, write the test charge with its sign, and distinguish force from energy change. “Charges move to lower potential” is incomplete. A charge released from rest under only the electric force initially moves toward lower electric potential energy; which direction that means depends on its sign.

Check your reasoning

Choose an answer before opening its feedback. Each option explains the reasoning behind that choice.

Check 1

Two fixed point sources have equal magnitudes: +Q is left of a midpoint and −Q is equally far to its right. Use zero potential at infinity. At the midpoint, which field/potential combination is correct?

A. Field left; potential zero.

Review this choice. The potentials cancel, but both field contributions point right: away from+Q and toward−Q.

B. Field left; potential positive.

Review this choice. Both the field direction and scalar sum are wrong. Equal-and-opposite source potentials cancel.

C. Field right; potential zero.

Correct. The two rightward field vectors add. The potentials kQ/r and−kQ/r cancel because potential is a signed scalar.

D. Field right; potential positive.

Review this choice. The field direction is correct, but equal-distance positive and negative contributions make the potential zero, not positive.

The principle to carry forward

A zero scalar potential need not mean a zero field.

Check 2

A +3.0 nC test charge is taken from a30 V point to a90 V point through a fixed electrostatic field. One route is straight and another is longer and curved. What work does the electric field do on each route?

A. −180 nJ on each route.

Correct. ΔV=+60 V, so electric work=−qΔV=−180 nJ. Fixed electrostatic sources make this work depend only on endpoints.

B. −180 nJ on the straight route and a more negative value on the longer route.

Review this choice. Length alone does not increase total electrostatic work. Contributions along the detour sum to the same endpoint potential difference.

C. +180 nJ on each route.

Review this choice. This is the sign of the potential-energy increase, not work by the electric field. An external agent can supply positive work for a slow transfer.

D. Zero on each route.

Review this choice. The endpoints differ in potential. A nonzero charge therefore has a nonzero potential-energy change and nonzero electric work.

The principle to carry forward

Name which agent’s work is requested before assigning its sign.

Check 3

Two large stationary parallel plates have the left plate at −40 V and the right plate at +80 V. The field between them is uniform. A negative test charge is released from rest midway between the plates; ignore gravity and all other forces. What initially happens?

A. It accelerates left, toward lower potential and lower potential energy.

Review this choice. The field points left toward lower potential, but the force on a negative charge points opposite the field, to the right.

B. It accelerates left, toward lower potential and higher potential energy.

Review this choice. An imposed leftward displacement would raise this negative charge’s potential energy, but the electric force drives it right.

C. It accelerates right, toward higher potential and lower potential energy.

Correct. Potential increases to the right, so the field points left. The negative charge’s force points right; for q < 0, increasing V decreases U.

D. It accelerates right, toward higher potential and higher potential energy.

Review this choice. The initial direction is correct. However, qΔV is negative for this negative charge moving toward higher potential, so its potential energy decreases.

The principle to carry forward

Determine the field from the supplied potential ordering, then apply the test-charge sign to force and potential energy.