Drill – potential at the midpoint

Drill: potential of two charges. Two point charges $q_1 = +3\ \mu\text{C}$ and $q_2 = -2\ \mu\text{C}$ are separated by $d = 1{,}0\ \text{m}$. Calculate the electric potential at the midpoint between the two charges.

Solution

Setup. Potential is a scalar: at the midpoint the contributions of the two charges add algebraically (superposition principle), each at distance $r = d/2 = 0{,}50$ m. $V = \frac{k q_1}{r} + \frac{k q_2}{r} = \frac{k}{r}\,(q_1 + q_2)$ Data. $q_1 = +3\cdot 10^{-6}$ C, $q_2 = -2\cdot 10^{-6}$ C, $r = 0{,}50$ m, $k = 9\cdot 10^9$. Substitution. $V = \frac{9\cdot 10^9}{0{,}50}\,\big(3\cdot 10^{-6} - 2\cdot 10^{-6}\big) = (1{,}8\cdot 10^{10})(1\cdot 10^{-6})$ Result. $\ev{V \approx 1{,}8\cdot 10^4 \ \text{V} = 18 \ \text{kV}}$ Note: the signs of the charges enter directly into the sum, unlike the field (a vector) where the directions would also matter.

Links

Topics: Electric field and potential Concepts: Electric potential · Electric charge Skills: Superposition principle Methods: Gradient method Objects: Charged sphere