A student maps the potential around a charge distribution and concludes that the electric field at every point must point toward the region of highest potential, since 'field lines go toward strong sources.' What is wrong with this reasoning?
ANothing — the electric field always points toward high potential
BThe electric field points in the direction of steepest decrease in potential (away from high V, toward low V) — the negative sign in E = −∇V reverses the gradient direction
CThe electric field is perpendicular to the potential gradient and has no component toward high or low potential
DThe electric field direction depends only on the sign of the source charge, not on the potential
The negative sign in E⃗ = −∇V is the critical physics. The gradient ∇V points in the direction of steepest increase of V, but the field is the negative gradient — it points toward lower potential. A positive test charge spontaneously accelerates from high V to low V, gaining kinetic energy as it loses potential energy, just as a ball rolls downhill. The field points 'downhill' on the potential surface, not uphill.
Question 2 Multiple Choice
On a field diagram, an equipotential line is drawn through a region. What must be true about electric field lines in that region?
AElectric field lines run parallel to the equipotential line, since they trace the same potential
BElectric field lines are perpendicular to the equipotential line, crossing it at right angles
CThere are no electric field lines where an equipotential exists
DElectric field lines and equipotential lines are the same thing labeled differently
Since E⃗ = −∇V and the gradient of V is always perpendicular to the level surfaces of V (the equipotentials), the field must be perpendicular to the equipotentials. If the field had any component along an equipotential surface, it would do work on a charge moving along that surface — but a charge moving along an equipotential does zero work by definition (ΔV = 0). Therefore the field can only be perpendicular to equipotential surfaces, never parallel.
Question 3 True / False
If the electric potential is constant throughout a region of space, the electric field in that region is zero.
TTrue
FFalse
Answer: True
E⃗ = −∇V. If V is constant, its gradient is zero in all directions, so E⃗ = 0. This is exactly what happens inside an ideal conductor at electrostatic equilibrium: the entire conductor (interior and surface) is at the same potential, so the internal field vanishes. Any non-zero field inside would accelerate free charges until they redistributed to eliminate the field — which is why equilibrium requires E = 0 inside.
Question 4 True / False
A positive test charge placed in an electric field will naturally accelerate from regions of low electric potential toward regions of high electric potential.
TTrue
FFalse
Answer: False
A positive charge accelerates from high potential to low potential. The electric force on a positive charge is F⃗ = qE⃗ = q(−∇V), pointing in the direction of decreasing V. This is analogous to gravity: a positive potential energy corresponds to a high position, and objects spontaneously move to lower potential energy. Moving from high V to low V decreases the electric potential energy (U = qV for positive q), so the charge gains kinetic energy — the motion is spontaneous in that direction, not the reverse.
Question 5 Short Answer
A positive charge is released from rest in a non-uniform electric field. Describe the relationship between the potential at its starting point and the direction it moves. Why does the negative sign in E⃗ = −∇V matter physically, not just mathematically?
Think about your answer, then reveal below.
Model answer: The charge accelerates toward lower potential. The negative sign means the field points in the direction of decreasing V — down the potential gradient, like a ball rolling downhill. Physically, the sign encodes the fact that a positive charge loses potential energy (U = qV) as it moves to lower V, and this loss becomes kinetic energy. Without the negative sign, the math would say the field points up the potential hill, which would predict spontaneous acceleration toward higher potential energy — violating conservation of energy.
Students often treat E⃗ = −∇V as a formula to memorize without grasping why the sign is there. The physical content is: positive charges roll downhill in potential, just as masses roll downhill in gravitational potential. The gradient ∇V points uphill; the field (−∇V) points downhill. This is why the sign matters: it determines the direction of force and therefore the direction of motion, not just the magnitude of the field.