A charge of 4 μ C is to be divided into two. The distance between the two divided charges is constant. The magnitude of the divided charges so that the force between them is maximum, will be:

Question

A charge of 4 &#956; C is to be divided into two. The distance between the two divided charges is constant. The magnitude of the divided charges so that the force between them is maximum, will be:

Aicharya · Accepted Answer

If after division, one of the charge is $q$ the other charge will be $(4 - q)$ . The force between them can be written as,
$F = \frac{K q (4 - q)}{d^{2}}$
For the force between them to be maximum,
$\frac{d F}{d q} = 0 \Rightarrow \frac{K}{d^{2}} [4 - 2 q] = 0$
$\Rightarrow q = 2 μC$

A charge of 4 μ C is to be divided into two. The distance between the two divided charges is constant.…

Solution

A charge of 4 &#956; C is to be divided into two. The distance between the two divided charges is constant.…

Solution

A charge of 4 μ C is to be divided into two. The distance between the two divided charges is constant.…