Potential difference and work

To move a charge $q = +5$ nC from point A to point B in the field, work $W = +4{,}5\cdot 10^{-7}$ J is done (positive, done against the field forces). What is the potential difference $V_B - V_A$? In which direction is $V$ greater?

Solution

Setup. The work done by an external agent against the field to move the charge from A to B is related to the potential difference by $W_\text{est} = q\,(V_B - V_A)$ Data. $q = +5 \text{ nC} = 5\cdot 10^{-9}$ C, $W_\text{est} = +4{,}5\cdot 10^{-7}$ J. Substitution. $V_B - V_A = \frac{W_\text{est}}{q} = \frac{4{,}5\cdot 10^{-7}}{5\cdot 10^{-9}}$ Result. $\ev{V_B - V_A = 90 \ \text{V}}$ Since $V_B - V_A > 0$, the potential is higher at B: indeed positive work is needed to bring a positive charge towards a higher potential, exactly like lifting a mass to greater heights in the gravitational field.

Links

Topics: Electric field and potential Concepts: Electric potential · Electrostatic potential energy Skills: Conservation of energy Methods: Gradient method Objects: Charged sphere