There is one long non-conducting solid cylinder of radius R. The volume charge density is given by `rho=kx^2` where k is a constant and x is the distance of the point from centre of cylinder. Electric field intensity inside the cylinder at a distance x from the centre is

Options

A. `(kx^2)/(3epsilon_0)`
B. `(kx^3)/(4epsilon_0)`
C. `(kx^3)/(2epsilon_0)`
D. `(kx^3)/epsilon_0`

Explanation

To find the electric field intensity inside a long non-conducting solid cylinder with a volume charge density given by \(\rho = kx^2\), where \(k\) is a constant and \(x\) is the distance from the center of the cylinder, we can follow these steps:

### Step 1: Define the Problem
We are given a long non-conducting solid cylinder of radius \(R\) and a volume charge density \(\rho = kx^2\). We need to find the electric field intensity \(E\) at a distance \(x\) from the center of the cylinder.

### Step 2: Choose a Gaussian Surface
To apply Gauss's law, we choose a cylindrical Gaussian surface of radius \(x\) and length \(L\) (where \(x < R\)). The electric field \(E\) will be uniform over the curved surface of the cylinder due to symmetry.

### Step 3: Calculate the Enclosed Charge
The charge density at a distance \(r\) from the center is given by:
\[ \rho(r) = kr^2 \]
To find the total charge \(dq\) enclosed within the Gaussian surface, we consider a thin cylindrical shell of radius \(r\) and thickness \(dr\):
\[ dq = \rho(r) \cdot dV \]
where \(dV\) is the volume of the thin shell:
\[ dV = 2\pi r \cdot dr \cdot L \]
Thus,
\[ dq = kr^2 \cdot (2\pi r \cdot dr \cdot L) = 2\pi k r^3 dr \cdot L \]
To find the total charge \(Q\) enclosed from \(0\) to \(x\), we integrate:
\[ Q = \int_0^x dq = \int_0^x 2\pi k r^3 L \, dr \]
Calculating the integral:
\[ Q = 2\pi k L \int_0^x r^3 \, dr = 2\pi k L \left[ \frac{r^4}{4} \right]_0^x = 2\pi k L \cdot \frac{x^4}{4} = \frac{\pi k L x^4}{2} \]

### Step 4: Apply Gauss's Law
According to Gauss's law:
\[ \Phi_E = \frac{Q}{\epsilon_0} \]
The electric flux \(\Phi_E\) through the Gaussian surface is given by:
\[ \Phi_E = E \cdot (2\pi x L) \]
Setting the two expressions for flux equal gives:
\[ E \cdot (2\pi x L) = \frac{\frac{\pi k L x^4}{2}}{\epsilon_0} \]

### Step 5: Solve for the Electric Field \(E\)
Now, we can solve for \(E\):
\[ E = \frac{\frac{\pi k L x^4}{2}}{\epsilon_0 \cdot (2\pi x L)} \]
Simplifying:
\[ E = \frac{k x^3}{4 \epsilon_0} \]

### Final Result
The electric field intensity inside the cylinder at a distance \(x\) from the center is:
\[ E = \frac{k x^3}{4 \epsilon_0} \]