Question

Two equal positive point charges q are held at a fixed distance a apart. A point test charge is located in a
plane that is normal to the line joining these charges and midway between them. Radius of the circle in
this plane for which the force on the test particle is maximum, will be
1. a/√2
2. a/2√2
3. a/2
4. √2a​

Answer 1

Given that,

Two equal positive point charges q are held at a fixed distance a apart.

Radius = r

Distance = a

The force due to single charge is

$F=\dfrac{1}{4\pi\epsilon_{0}}\times\dfrac{q}{(r^2+a^)}$

We need to calculate the electric field

Using formula of electric field

$E=2F_{1}\cos(90-\theta$

$E=2\times\dfrac{1}{4\pi\epsilon_{0}}\dfrac{q}{(r^2+a^2)}\cos(90-\theta)$

$E=2\times\dfrac{1}{4\pi\epsilon_{0}}\times\dfrac{q}{(r^2+a^2)}\sin\theta$

$E=2\times\dfrac{1}{4\pi\epsilon_{0}}\times\dfrac{q}{(r^2+a^2)}\dfrac{r}{\sqrt{r^2+a^2}}$

$E=\dfrac{2}{4\pi\epsilon_{0}}\times\dfrac{r}{(r^2+a^2)^{\frac{3}{2}}}$...(I)

We need to calculate the radius of the circle

Using equation (I)

$E=\dfrac{2}{4\pi\epsilon_{0}}\times\dfrac{r}{(r^2+a^2)^{\frac{3}{2}}}$

On differentiating

$\dfrac{dE}{dt}=\dfrac{2q}{4\pi\epsilon_{0}}\times\dfrac{(r^2+a^2)^{\frac{3}{2}}\times l+r(\frac{3}{2}(a^2+r^2)^{\frac{1}{2}}\times2r)}{(r^2+a^2)^3}$

For electric field to be maximum

When, $\dfrac{dE}{dt}=0$

$\dfrac{2q}{4\pi\epsilon_{0}}\times\dfrac{(r^2+a^2)^{\frac{3}{2}}\times l+r(\frac{3}{2}(a^2+r^2)^{\frac{1}{2}}\times2r)}{(r^2+a^2)^3}=0$

$(a^2+r^2)^{\frac{1}{2}}=\dfrac{2\times3r^2}{2}\times(a^2+r^2)^{\frac{1}{2}}$

$r=\sqrt{2}(a)$

Hence, The radius of the circle is $\sqrt{2}(a)$

(4) is correct option.