Question 803: Electrostats & Capacitance

Question

Consider a long uniformly charged cylinder having constant volume charge density ' $\lambda$ ' and radius ' $R$ '. A Gaussian surface is in the form of a cylinder of radius ' $r$ ' such that vertical axis of both the cylinders coincide. For a point inside the cylinder $(r< R)$, electric field is directly proportional to

Accepted Answer

To determine how the electric field varies with the radial distance inside the cylinder, we can use Gauss's law. HereÃ¢ÂÂ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:
$$	ext{Flux} = E \cdot (2\pi r L)$$
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 $$ho$$), the charge inside the Gaussian surface is:
$$Q_{	ext{enc}} = ho \cdot 	ext{Volume of the Gaussian cylinder} = ho \cdot (\pi r^2 L)$$
(Note: Sometimes $$\lambda$$ is used to denote linear charge density, but here we assume the cylinder is uniformly charged in volume and use $$ho$$.)
Applying Gauss's Law:
Gauss's law states:
$$\oint \vec{E} \cdot d\vec{A} = \frac{Q_{	ext{enc}}}{\epsilon_0}$$
Substituting the above expressions:
$$E \cdot (2\pi r L) = \frac{ho \cdot (\pi r^2 L)}{\epsilon_0}$$
Solving for the Electric Field $$E$$:
Cancel common factors $$\pi$$ and $$L$$ on both sides:
$$E \cdot (2r) = \frac{ho \cdot r^2}{\epsilon_0}$$
Rearranging gives:
$$E = \frac{ho \, r}{2\epsilon_0}$$
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: $$r$$.

Answer

$\mathrm{r}^{-1}$

Answer

$\mathrm{r}$

Answer

$\mathrm{r}^2$

Answer

$\mathrm{r}^{-2}$

Electrostats & Capacitance

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