Question

Electric field at point P is given by $E=rE_0$. The total flux through the given cylinder of radius R and height h is

• A. E_0 \pi R^2 h
• B. 2 E_0 \pi R^2 h
• C. 3 E_0 \pi R^2 h
• D. 4 E_0 \pi R^2 h

Answer 1

Answer: (B)

2 E_0 \pi R^2 h

Answer 2

The electric field is given by $E = rE_0$. Assuming this means the electric field is radial and its magnitude depends on the radial distance $r$ from the central axis of the cylinder, we can write the electric field vector as $\vec{E} = E_0 r \hat{r}$, where $\hat{r}$ is the radial unit vector.

To find the total flux through the cylinder, we can use Gauss's divergence theorem: $\Phi = \oint_S \vec{E} \cdot d\vec{A} = \int_V (\nabla \cdot \vec{E}) dV$

First, we calculate the divergence of the electric field in cylindrical coordinates. For $\vec{E} = E_0 r \hat{r}$, the components are $E_r = E_0 r$, $E_\theta = 0$, and $E_z = 0$. The divergence in cylindrical coordinates is: $\nabla \cdot \vec{E} = \frac{1}{r} \frac{\partial}{\partial r} (r E_r) + \frac{1}{r} \frac{\partial E_\theta}{\partial \theta} + \frac{\partial E_z}{\partial z}$ Substituting the components: $\nabla \cdot \vec{E} = \frac{1}{r} \frac{\partial}{\partial r} (r (E_0 r)) + \frac{1}{r} \frac{\partial (0)}{\partial \theta} + \frac{\partial (0)}{\partial z} = \frac{1}{r} \frac{\partial}{\partial r} (E_0 r^2) = \frac{1}{r} (2 E_0 r) = 2 E_0$ The divergence of the electric field is a constant, $2E_0$.

Now, we integrate the divergence over the volume of the cylinder. The volume of the cylinder of radius $R$ and height $h$ is $V = \pi R^2 h$. $\Phi = \int_V (\nabla \cdot \vec{E}) dV = \int_V (2 E_0) dV = 2 E_0 \int_V dV$ $\Phi = 2 E_0 V = 2 E_0 (\pi R^2 h)$

Alternatively, we can calculate the flux through each surface of the cylinder:

1. Top surface: The area vector is $d\vec{A}_{top} = dA \hat{k}$. Since $\vec{E} = E_0 r \hat{r}$, $\vec{E} \cdot d\vec{A}_{top} = (E_0 r \hat{r}) \cdot (dA \hat{k}) = 0$. So, $\Phi_{top} = 0$.
2. Bottom surface: The area vector is $d\vec{A}_{bottom} = -dA \hat{k}$. $\vec{E} \cdot d\vec{A}_{bottom} = (E_0 r \hat{r}) \cdot (-dA \hat{k}) = 0$. So, $\Phi_{bottom} = 0$.
3. Curved lateral surface: The radius is $r=R$. The area vector is $d\vec{A}_{lateral} = R d\theta dz \hat{r}$. The electric field on this surface is $\vec{E} = E_0 R \hat{r}$. $\vec{E} \cdot d\vec{A}_{lateral} = (E_0 R \hat{r}) \cdot (R d\theta dz \hat{r}) = E_0 R^2 d\theta dz$ Integrating over the curved surface: $\Phi_{lateral} = \int_{z=0}^{h} \int_{\theta=0}^{2\pi} E_0 R^2 d\theta dz = E_0 R^2 \int_{0}^{h} dz \int_{0}^{2\pi} d\theta = E_0 R^2 (h)(2\pi) = 2\pi R^2 h E_0$ The total flux is the sum of fluxes through all surfaces: $\Phi_{total} = \Phi_{top} + \Phi_{bottom} + \Phi_{lateral} = 0 + 0 + 2\pi R^2 h E_0 = 2 E_0 \pi R^2 h$.

Both methods yield the same result.