A solid sphere of radius `R` has total charge `2Q` and volume charge density `p=kr` where `r` is distance from centre. Now charges `Q` and `-Q` are placed diametrically opposite at distance `2a` where a is distance form centre of sphere such that net force on charge `Q` is zero then relation between a and `R` is

Options

A. `a= 8^(-1//4)R`
B. `a = (3R)/(2^¼)`
C. `a = 2^(-1//4) R`
D. `a = R//sqrt(3)`

Answer

To solve the problem, we need to find the relationship between the distance \( a \) from the center of the sphere to the charges and the radius \( R \) of the sphere, given that the net force on charge \( Q \) is zero.

### Step-by-Step Solution:

1. **Understanding the Charge Distribution**:
The sphere has a total charge of \( 2Q \) and a volume charge density \( \rho = kr \). The charge density varies linearly with the distance \( r \) from the center of the sphere.

2. **Finding the Electric Field Inside the Sphere**:
We can use Gauss's law to find the electric field \( E \) at a distance \( r \) from the center. The total charge enclosed within a Gaussian surface of radius \( r \) is given by:
\[ Q_{\text{enc}} = \int_0^r \rho \, dV = \int_0^r kr \cdot 4\pi r^2 \, dr = 4\pi k \int_0^r r^3 \, dr = 4\pi k \left[ \frac{r^4}{4} \right]_0^r = \pi k r^4 \]
By Gauss's law:
\[ \oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\epsilon_0} \implies E \cdot 4\pi r^2 = \frac{\pi k r^4}{\epsilon_0} \]
Thus, the electric field \( E \) inside the sphere is:
\[ E = \frac{k r^2}{4 \epsilon_0} \]

3. **Finding the Value of \( k \)**:
The total charge \( Q_{\text{total}} = 2Q \) can also be expressed as:
\[ Q_{\text{total}} = \int_0^R \rho \, dV = \int_0^R kr \cdot 4\pi r^2 \, dr = 4\pi k \int_0^R r^3 \, dr = 4\pi k \left[ \frac{R^4}{4} \right]_0^R = \pi k R^4 \]
Setting this equal to \( 2Q \):
\[ \pi k R^4 = 2Q \implies k = \frac{2Q}{\pi R^4} \]

4. **Substituting \( k \) into the Electric Field**:
Now substituting \( k \) back into the expression for \( E \):
\[ E = \frac{2Q}{\pi R^4} \cdot \frac{r^2}{4 \epsilon_0} = \frac{Q r^2}{2 \pi \epsilon_0 R^4} \]

5. **Calculating the Forces on Charge \( Q \)**:
The force \( F_{AS} \) on charge \( Q \) due to the electric field at distance \( a \) is:
\[ F_{AS} = E \cdot Q = \frac{Q a^2}{2 \pi \epsilon_0 R^4} \cdot Q = \frac{Q^2 a^2}{2 \pi \epsilon_0 R^4} \]
The force \( F_{AB} \) due to charge \( -Q \) placed at distance \( 2a \) is given by Coulomb's law:
\[ F_{AB} = \frac{1}{4 \pi \epsilon_0} \cdot \frac{Q^2}{(2a)^2} = \frac{Q^2}{16 \pi \epsilon_0 a^2} \]

6. **Setting the Forces Equal**:
For the net force on charge \( Q \) to be zero:
\[ F_{AS} = F_{AB} \implies \frac{Q^2 a^2}{2 \pi \epsilon_0 R^4} = \frac{Q^2}{16 \pi \epsilon_0 a^2} \]
Canceling \( Q^2 \) and \( \pi \epsilon_0 \) from both sides:
\[ \frac{a^2}{2 R^4} = \frac{1}{16 a^2} \implies 16 a^4 = 2 R^4 \implies 8 a^4 = R^4 \implies a = \frac{R}{\sqrt[4]{8}} = \frac{R}{2^{3/4}} \]

### Final Relation:
Thus, the relation between \( a \) and \( R \) is:
\[ a = \frac{R}{2^{3/4}} \]