What is the work done by the electric field in moving a test charge q from point a to b with potential difference ΔV?

What is the work done by the electric field in moving a test charge q from point a to b with potential difference ΔV?

• A. W = ΔV / q
• B. W = ∫ q E · dl
• C. W = q ΔV
• D. W = − q ΔV

Answer: (C)

The work done by the electric field is tied to how the charge’s potential energy changes as it moves. For a charge q, potential energy is U = qV. Moving from point a to point b changes the potential energy by ΔU = q[V(b) − V(a)]. The work done by the field equals the negative of that change: W_field = −ΔU = − q [V(b) − V(a)]. Whether this looks like a negative or positive multiple of ΔV depends on how ΔV is defined. If ΔV is taken as V(b) − V(a), then W_field = − q ΔV. If, however, ΔV is defined with the opposite sign, ΔV = V(a) − V(b), then W_field becomes W_field = q ΔV. In this question, ΔV is defined as V(a) − V(b), so the work done by the field is W = q ΔV. This also makes intuitive sense: if starting at a higher potential and moving to a lower potential for a positive charge, the field does positive work on the charge.

The work done by the electric field is tied to how the charge’s potential energy changes as it moves. For a charge q, potential energy is U = qV. Moving from point a to point b changes the potential energy by ΔU = q[V(b) − V(a)]. The work done by the field equals the negative of that change: W_field = −ΔU = − q [V(b) − V(a)].

Whether this looks like a negative or positive multiple of ΔV depends on how ΔV is defined. If ΔV is taken as V(b) − V(a), then W_field = − q ΔV. If, however, ΔV is defined with the opposite sign, ΔV = V(a) − V(b), then W_field becomes W_field = q ΔV.

In this question, ΔV is defined as V(a) − V(b), so the work done by the field is W = q ΔV. This also makes intuitive sense: if starting at a higher potential and moving to a lower potential for a positive charge, the field does positive work on the charge.