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

Correct Answer: 2

To find the point on the equatorial line where the force on a test charge \( q_0 \) is maximum, consider the setup with two equal positive point charges \( +Q \) separated by a distance \( 2a \). The equatorial line is perpendicular to the line joining the charges and passes through their midpoint.

The force \( F \) on a test charge \( q_0 \) at a distance \( y \) from the midpoint along the equatorial line due to one charge \( +Q \) is given by \( F = \frac{kQq_0}{r^2} \), where \( k \) is Coulomb's constant and \( r \) is the distance from the charge to the point \( y \). By geometry, \( r = \sqrt{a^2 + y^2} \).

The net force \( F_{\text{net}} \) due to both charges is along the y-axis, as the x-components cancel out. Thus, it's given by \( F_{\text{net}} = 2F \cos\theta \), where \( \cos\theta = \frac{a}{\sqrt{a^2+y^2}} \). Thus:

\( F_{\text{net}} = 2 \left(\frac{kQq_0}{a^2+y^2}\right) \left(\frac{a}{\sqrt{a^2+y^2}}\right) = \frac{2kQq_0a}{(a^2+y^2)^{3/2}} \)

To maximize \( F_{\text{net}} \), differentiate w.r.t. \( y \) and set the derivative to zero:

\(\frac{d}{dy}\left(\frac{2kQq_0a}{(a^2+y^2)^{3/2}}\right) = 0\)

After simplification, we find \( y = \frac{a}{\sqrt{2}} \).

So, the value of \( x \) such that \( \frac{a}{\sqrt{x}} = \frac{a}{\sqrt{2}} \) is \( x = 2 \). This value fits within the given range (2,2).