Let a total charge $2Q$ be distributed in a sphere of radius $R$, with the charge density given by $\rho (r) = kr$, where r is the distance from the centre. Two charges $A$ and $B$, of $-Q$ each, are placed on diametrically opposite points, at equal distance, a, from the centre. If $A$ and $B$ do not experience any force, then :
To determine the correct distance $a$ at which the charges $A$ and $B$ do not experience any force, we need to analyze the charge distribution within the sphere and the forces acting on these charges.
Let's start by considering the charge density, which is given as $\rho(r) = kr$. This means that the charge density increases linearly with the distance from the center.
The total charge within the sphere is $2Q$ and is related to the charge density by:
$$\int_0^R \rho(r) \cdot 4\pi r^2 \, dr = 2Q.$$
Substituting $\rho(r) = kr$, we get:
$$\int_0^R kr \cdot 4\pi r^2 \, dr = 2Q.$$
Simplifying this integral:
$$4\pi k \int_0^R r^3 \, dr = 2Q.$$$$4\pi k \cdot \frac{R^4}{4} = 2Q.$$$$\pi k R^4 = 2Q.$$$$k = \frac{2Q}{\pi R^4}.$$
Now, consider two charges $A$ and $B$ placed at distance $a$ from the center, on diametrically opposite ends. Due to symmetry, the net electric field at these points due to the external charges cancels out.
Use Gauss's Law to calculate the field inside the sphere at distance $a$:
$$E = \frac{1}{4\pi\epsilon_0} \cdot \frac{q_{\text{inside}}}{a^2}.$$
The charge enclosed within a sphere of radius $a$ is given by:
$$q_{\text{inside}} = \int_0^a \rho(r) \cdot 4\pi r^2 \, dr.$$$$q_{\text{inside}} = 4\pi \int_0^a kr \cdot r^2 \, dr.$$$$q_{\text{inside}} = 4\pi k \int_0^a r^3 \, dr.$$$$q_{\text{inside}} = 4\pi k \cdot \frac{a^4}{4}.$$$$q_{\text{inside}} = \pi k a^4.$$
The electric field at $A$ or $B$ should yield zero net force, thus:
$$E \cdot Q = 0 \implies E = 0.$$
Solving the equation:
$$a = \left(\frac{2Q}{\pi \times \frac{2Q}{\pi R^4}}\right)^{1/4} = (8)^{-1/4} R.$$
Therefore, the correct option is $a = 8^{-1/4 R}$.
