Source: High school physics (Chinese)
Problem Sets:
Problem
As shown, $AB = 2l$. The arc $OCD$ is a semicircle of radius $l$ centered at $B$, in the plane of the page. Point $A$ carries charge $+q$ and point $B$ carries charge $-q$. Points on the $x$-axis are at $A = (0,0)$, $O = (l, 0)$, $B = (2l, 0)$, $D = (3l, 0)$, and $C = (2l, l)$ is the top of the semicircle.
- Find the work done by the electric field on a unit positive charge as it is moved from $O$ along the semicircular arc $OCD$ to $D$.
- Find the work done by the electric field on a unit negative charge as it is moved from $D$ along the extension of line $AB$ to infinity.
(1) $W_1 = \dfrac{2kq}{3l}$. (2) $W_2 = \dfrac{2kq}{3l}$ (same magnitude as (1)).
The work done by the electric field on a charge $q'$ moved from initial point $i$ to final point $f$ depends only on the potentials: $W = q'(V_i - V_f)$.
Potentials (take $V = 0$ at infinity, $V = \tfrac{kq}{r}$ from each source):
At $O$ (distance $l$ from $A$ and distance $l$ from $B$):
$$V_O = \dfrac{kq}{l} + \dfrac{k(-q)}{l} = 0.$$At $D$ (distance $3l$ from $A$ and distance $l$ from $B$):
$$V_D = \dfrac{kq}{3l} + \dfrac{k(-q)}{l} = -\dfrac{2kq}{3l}.$$(1) Unit positive charge from $O$ to $D$:
$$W_1 = (+1)(V_O - V_D) = 0 - \left(-\dfrac{2kq}{3l}\right) = \dfrac{2kq}{3l}.$$(2) Unit negative charge from $D$ to infinity ($V_\infty = 0$):
$$W_2 = (-1)(V_D - V_\infty) = (-1)\left(-\dfrac{2kq}{3l}\right) = \dfrac{2kq}{3l}.$$