Question

In the given figure flux through surface $S_1$ is $\phi_1$ & through $S_2$ is $\phi_2$. Which is correct?

• A. $\phi_1 = \phi_2$
• B. $\phi_1 > \phi_2$
• C. $\phi_1 < \phi_2$
• D. None of these

Answer 1

Answer: (A)

$\phi_1 = \phi_2$

Answer 2

The problem asks to compare the electric flux through two surfaces, $S_1$ and $S_2$, due to a point charge 'q'. Both surfaces are shown as cross-sections of the same cone originating from the point charge 'q'.

According to the definition of electric flux for a point charge 'q' through an open surface, the flux $\Phi$ is given by: $\Phi = \int \vec{E} \cdot d\vec{A}$ where $\vec{E}$ is the electric field and $d\vec{A}$ is the area vector. For a point charge, the electric field is radial and its magnitude is $E = \frac{q}{4\pi\epsilon_0 r^2}$. Substituting this into the integral: $\Phi = \int \frac{q}{4\pi\epsilon_0 r^2} \hat{r} \cdot d\vec{A}$ where $\hat{r}$ is the unit vector pointing radially outward from 'q'. The term $\frac{\hat{r} \cdot d\vec{A}}{r^2}$ is the differential solid angle $d\Omega$ subtended by the area element $d\vec{A}$ at the point charge 'q'. Therefore, the total electric flux through the surface is: $\Phi = \frac{q}{4\pi\epsilon_0} \int d\Omega = \frac{q \Omega}{4\pi\epsilon_0}$ This formula shows that the electric flux through any surface due to a point charge is directly proportional to the solid angle $\Omega$ subtended by that surface at the point charge.

In the given figure, both surfaces $S_1$ and $S_2$ are shown as cross-sections of the same cone originating from the point charge 'q'. This implies that both surfaces subtend the same solid angle $\Omega$ at the point charge 'q'.

Since $\phi_1 = \frac{q \Omega_1}{4\pi\epsilon_0}$ and $\phi_2 = \frac{q \Omega_2}{4\pi\epsilon_0}$, and from the diagram, $\Omega_1 = \Omega_2 = \Omega$, it follows that: $\phi_1 = \phi_2$ This is a fundamental concept in electrostatics: the number of electric field lines passing through any surface that intercepts the same bundle of field lines from a point charge is constant, provided there are no charges between the surfaces. Electric flux is proportional to the number of field lines.

Therefore, the correct relationship is $\phi_1 = \phi_2$.

The provided solution in the image indicates $\phi_1 > \phi_2$ as correct, which contradicts the fundamental principle of electric flux for a point charge when surfaces subtend the same solid angle. Based on the standard understanding of electric flux and the visual representation, $\phi_1 = \phi_2$ is the correct answer.