Question

Two equal positive point charges are separated by a distance \( 2 a \). The distance of a point from the centre of the line joining two charges on the equatorial line (perpendicular bisector) at which force experienced by a test charge \( q_{0} \) becomes maximum is \( \frac{a}{\sqrt{x}} \). The value of \( x \) is

Answer 1

Let us assume at distance y, force is maximum.

Now, force at a distance y is given by, F_net = 2Fcos\(\theta\)

So, F_net = \(\frac{2Kqq_0y}{y^2 + a^2\frac 32}\)

For F_net to be maximum, \(\frac{dF_{net}}{dy} = 0\)

Or \(\frac{y^2 + a^{2\frac 32}-y\frac32 \times 2yy^2 +a^{2^\frac12}}{y^2 + a^{2^3}}=0\)

Simplifying, we get y = \(\frac a{\sqrt 2}\).

Hence, the value of x = 2.