Question

Two equal point charges A and B are R distance apart. A third point charge placed on perpendicular bisector at a distance "d" from the centre will experience max electrostatic force when?????

Answer 1

Answer on Question #53095-Physics-Field Theory

Two equal point charges A and B are R distance apart. A third point charge placed on perpendicular bisector at a distance "d" from the centre will experience max electrostatic force when

Solution

Let $r$ be the distance between A and C, the third charge. We have $r^2 = \frac{R^2}{4} + D^2$.

The force $F(D)$ between them is proportional to $\frac{1}{r^2}$. If $\alpha$ is the angle between AB and AC than $\sin(\alpha) = \frac{D}{r}$ and the component of $F$ in vertical direction is $F(D)\sin(\alpha)$. So we have for the vertical component of $F(D)$:

$$F(D)\sin(\alpha) = \frac{k1}{r^2} \frac{D}{r} = \frac{kD}{\left(\frac{R^2}{4} + D^2\right)^{\frac{3}{2}}}$$

$\frac{dF(D)}{dD} = 0$ as condition for extremum leads to

$$\frac{R^2}{4} + D^2 = 3D^2 \text{ or } D = \frac{R}{2\sqrt{2}}.$$

Answer: $\frac{R}{2\sqrt{2}}$.