The potential difference ΔV between two points is related to the electric field via ΔV = -∫ E · dl. What sign does this give for work done by the field?

The potential difference ΔV between two points is related to the electric field via ΔV = -∫ E · dl. What sign does this give for work done by the field?

• A. The field does negative work when moving a positive test charge along the field lines
• B. The field does zero work
• C. The field does positive work when moving a positive test charge in the field direction
• D. The field does positive work when moving opposite to the field

Answer: (C)

The sign of work by the field follows from how potential difference and the field relate to each other. By definition, V_final − V_initial = − ∫ from initial to final of E · dl. The work done by the electric field on a test charge q as it moves along that path is W = ∫ F · dl = ∫ (qE) · dl = q ∫ E · dl. Because ∫ E · dl = −ΔV, we have W = −q ΔV. When a positive charge moves in the direction of the field, E and dl point the same way, so E · dl is positive and the integral is positive. That makes ΔV negative (the potential drops along the field), and thus W = −q ΔV is positive. So the field does positive work on a positive test charge moving along the field direction. If you move opposite to the field, the work would be negative; if there’s no component along the field, the work is zero.

The sign of work by the field follows from how potential difference and the field relate to each other. By definition, V_final − V_initial = − ∫ from initial to final of E · dl. The work done by the electric field on a test charge q as it moves along that path is W = ∫ F · dl = ∫ (qE) · dl = q ∫ E · dl. Because ∫ E · dl = −ΔV, we have W = −q ΔV.

When a positive charge moves in the direction of the field, E and dl point the same way, so E · dl is positive and the integral is positive. That makes ΔV negative (the potential drops along the field), and thus W = −q ΔV is positive. So the field does positive work on a positive test charge moving along the field direction. If you move opposite to the field, the work would be negative; if there’s no component along the field, the work is zero.