Two point charges $P = +25\ \mu C$ and $Q = -16\ \mu C$ are placed 5 cm apart. Find the position of the point at which the resultant electric field is zero:

Let the point where the electric field is zero be at a distance $x$ from charge $Q$. Since the charges are of opposite signs, the null point lies outside the region between the charges, closer to the charge with the smaller magnitude ($Q$).

Distance of the point from $Q = x$
Distance of the point from $P = x + 5$

For the net electric field to be zero, the magnitudes of the electric fields due to both charges must be equal:

$\dfrac{k|P|}{(x+5)^2} = \dfrac{k|Q|}{x^2}$

Substituting the given values:

$\dfrac{k \times 25}{(x+5)^2} = \dfrac{k \times 16}{x^2}$

Taking the square root on both sides:

$\dfrac{5}{x+5} = \dfrac{4}{x}$

$5x = 4(x+5)$

$5x = 4x + 20$

$x = 20$ cm

The distance from $Q$ is $20$ cm and the distance from $P$ is $20 + 5 = 25$ cm.

Answer: 20cm from Q and 25cm from P on the dipole axis