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) `a=8^ -(1/4)R`
B. b) `a=R`
C. c) `a=2R`
D. d) `a=3R//4`

Answer

To solve the problem, we need to analyze the forces acting on the charges A and B placed inside a sphere with a specific charge distribution. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Charge Distribution
The charge density is given by \( \rho(r) = kr \), where \( r \) is the distance from the center of the sphere. The total charge distributed in the sphere is \( 2Q \).

### Step 2: Calculate the Total Charge
To find the constant \( k \), we need to calculate the total charge using the charge density:
\[ Q = \int_0^R \rho(r) \, dV \]
Where \( dV = 4\pi r^2 \, dr \). Thus,
\[ Q = \int_0^R kr \cdot 4\pi r^2 \, dr = 4\pi k \int_0^R r^3 \, dr = 4\pi k \left[ \frac{r^4}{4} \right]_0^R = \pi k R^4 \]
Setting this equal to \( 2Q \):
\[ \pi k R^4 = 2Q \implies k = \frac{2Q}{\pi R^4} \]

### Step 3: Calculate the Electric Field Inside the Sphere
Using Gauss's law, the electric field \( E \) at a distance \( r \) from the center of the sphere is given by:
\[ E \cdot 4\pi r^2 = \frac{Q_{\text{enc}}}{\epsilon_0} \]
Where \( Q_{\text{enc}} \) is the charge enclosed within radius \( r \):
\[ Q_{\text{enc}} = \int_0^r \rho(r') \, dV = \int_0^r k r' \cdot 4\pi r'^2 \, dr' = 4\pi k \int_0^r r'^3 \, dr' = 4\pi k \left[ \frac{r'^4}{4} \right]_0^r = \pi k r^4 \]
Thus,
\[ E \cdot 4\pi r^2 = \frac{\pi k r^4}{\epsilon_0} \implies E = \frac{k r^2}{4 \epsilon_0} \]

### Step 4: Substitute \( k \) into the Electric Field Expression
Substituting \( k = \frac{2Q}{\pi R^4} \) into the electric field expression:
\[ E = \frac{\frac{2Q}{\pi R^4} r^2}{4 \epsilon_0} = \frac{Q r^2}{2 \pi R^4 \epsilon_0} \]

### Step 5: Calculate the Force on Charges A and B
The force on charge A due to the electric field at distance \( a \) is:
\[ F_{AS} = qE = -Q \cdot \frac{Q a^2}{2 \pi R^4 \epsilon_0} \]
The force between charges A and B (which are at a distance \( 2a \)) is given by Coulomb's law:
\[ F_{AB} = \frac{1}{4\pi \epsilon_0} \cdot \frac{(-Q)(-Q)}{(2a)^2} = \frac{Q^2}{16\pi \epsilon_0 a^2} \]

### Step 6: Set Forces Equal
Since charges A and B do not experience any net force, we set \( F_{AS} = F_{AB} \):
\[ \frac{Q^2 a^2}{2 \pi R^4 \epsilon_0} = \frac{Q^2}{16\pi \epsilon_0 a^2} \]
Cancelling \( Q^2 \) and \( \pi \epsilon_0 \) from both sides gives:
\[ \frac{a^2}{2 R^4} = \frac{1}{16 a^2} \]
Cross-multiplying leads to:
\[ 16 a^4 = 2 R^4 \implies a^4 = \frac{R^4}{8} \implies a = \frac{R}{2^{1/4}} = \frac{R}{\sqrt[4]{8}} = \frac{R}{2^{3/4}} \]

### Final Answer
Thus, the distance \( a \) from the center to the charges A and B is:
\[ a = \frac{R}{2^{3/4}} \]