To determine how the electric field varies with the radial distance inside the cylinder, we can use Gauss's law. Here’s the step-by-step reasoning:

Gaussian Surface:

We choose a cylindrical Gaussian surface of radius $$r$$ (with $$r < R$$) and length $$L$$, aligned coaxially with the charged cylinder.

Electric Flux Through the Gaussian Surface:

The electric field $$\vec{E}$$ points radially outward.

The flux through the curved side is given by the product of the electric field and the lateral surface area:

$$\text{Flux} = E \cdot (2\pi r L)$$

The end caps do not contribute because the electric field is parallel to those surfaces.

Charge Enclosed by the Gaussian Surface:

If the cylinder has a constant volume charge density (commonly denoted by $$\rho$$), the charge inside the Gaussian surface is:

$$Q_{\text{enc}} = \rho \cdot \text{Volume of the Gaussian cylinder} = \rho \cdot (\pi r^2 L)$$

(Note: Sometimes $$\lambda$$ is used to denote linear charge density, but here we assume the cylinder is uniformly charged in volume and use $$\rho$$.)

Applying Gauss's Law:

Gauss's law states:

$$\oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\epsilon_0}$$

Substituting the above expressions:

$$E \cdot (2\pi r L) = \frac{\rho \cdot (\pi r^2 L)}{\epsilon_0}$$

Solving for the Electric Field $$E$$:

Cancel common factors $$\pi$$ and $$L$$ on both sides:

$$E \cdot (2r) = \frac{\rho \cdot r^2}{\epsilon_0}$$

Rearranging gives:

$$E = \frac{\rho \, r}{2\epsilon_0}$$

This shows that the electric field inside the cylinder is directly proportional to $$r$$.

Thus, for a point inside the cylinder ($$r < R$$), the electric field is directly proportional to $$r$$.

The correct answer is Option B: $$r$$.

Electrostatics

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