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 charge 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:

Options

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

Explanation

To solve the problem, we need to analyze the situation step by step, considering the distribution of charge and the forces acting on charges A and B.

### Step 1: Understand the Charge Distribution
We have a total charge of \(2Q\) distributed in a sphere of radius \(R\) with a charge density given by:
\[ \rho(r) = kr \]
where \(r\) is the distance from the center of the sphere.

### Step 2: Calculate the Total Charge
To find the value of \(k\), we need to ensure that the total charge within the sphere equals \(2Q\). The total charge can be calculated by integrating the charge density over the volume of the sphere:
\[ Q_{\text{total}} = \int_0^R \rho(r) \, dV = \int_0^R kr \cdot 4\pi r^2 \, dr \]
This gives:
\[ Q_{\text{total}} = 4\pi k \int_0^R r^3 \, dr = 4\pi k \left[\frac{r^4}{4}\right]_0^R = \pi k R^4 \]
Setting this equal to \(2Q\):
\[ \pi k R^4 = 2Q \implies k = \frac{2Q}{\pi R^4} \]

### Step 3: Calculate the Electric Field Inside the Sphere
Using Gauss's law, we can find the electric field \(E\) at a distance \(a\) from the center (where charges A and B are placed). The charge enclosed by a Gaussian surface of radius \(a\) is:
\[ Q_{\text{enc}} = \int_0^a \rho(r) \, dV = \int_0^a kr \cdot 4\pi r^2 \, dr = 4\pi k \int_0^a r^3 \, dr = 4\pi k \left[\frac{r^4}{4}\right]_0^a = \pi k a^4 \]
Substituting \(k\):
\[ Q_{\text{enc}} = \pi \left(\frac{2Q}{\pi R^4}\right) a^4 = \frac{2Q a^4}{R^4} \]

Using Gauss's law:
\[ \oint E \cdot dA = \frac{Q_{\text{enc}}}{\epsilon_0} \implies E \cdot 4\pi a^2 = \frac{2Q a^4}{\epsilon_0 R^4} \]
Thus,
\[ E = \frac{2Q a^2}{4\pi \epsilon_0 R^4} = \frac{Q a^2}{2\pi \epsilon_0 R^4} \]

### Step 4: Calculate the Forces on Charges A and B
The force on charge A due to charge B is given by Coulomb's law:
\[ F_{AB} = \frac{1}{4\pi \epsilon_0} \frac{(-Q)(-Q)}{(2a)^2} = \frac{Q^2}{4\pi \epsilon_0 (2a)^2} = \frac{Q^2}{16\pi \epsilon_0 a^2} \]
The force on charge A due to the electric field from the sphere is:
\[ F_{ES} = Q \cdot E = Q \cdot \frac{Q a^2}{2\pi \epsilon_0 R^4} = \frac{Q^2 a^2}{2\pi \epsilon_0 R^4} \]

### Step 5: Set the Forces Equal
Since charges A and B do not experience any force, we set \(F_{AB} = F_{ES}\):
\[ \frac{Q^2}{16\pi \epsilon_0 a^2} = \frac{Q^2 a^2}{2\pi \epsilon_0 R^4} \]
Cancelling \(Q^2\) and rearranging gives:
\[ \frac{1}{16 a^2} = \frac{a^2}{2 R^4} \implies 2 R^4 = 16 a^4 \implies R^4 = 8 a^4 \implies a = \frac{R}{\sqrt[4]{8}} = \frac{R}{2^{3/4}} \]

### Final Answer
Thus, we find:
\[ a = \frac{R}{2^{3/4}} \]