Practicing Success
a b c d |
a |
$\text{Potential energy will be minimum when distance between 2q and 8q is maximum}$ $ \text{Charge q is placed between 2q and 8q}$ $\text{ Let q is placed at a distance x from charge 2q}$ $ U = k( \frac{2q^2}{x} + \frac{8q^2}{9-x} + \frac{16q^2}{9})$ $ \Rightarrow U = kq^2( \frac{2}{x} + \frac{8}{9-x} + \frac{16}{9})$ $\text{For U to be minimum } \frac{2}{x} + \frac{8}{9-x} \text{should be mimimun}$ $\frac{d}{dx}(\frac{2}{x} + \frac{8}{9-x}) = 0 $ $\Rightarrow x = 3 cm$
|