Question

A point charge $q$ is fixed at origin. The electric flux passing through infinite surface at $x=a$ and parallel to $y-z$ plane is $\frac{q}{n\epsilon_0}$. Find $n$.

Answer 1

2

Answer 2

The problem asks us to find the value of $n$ if the electric flux passing through an infinite surface at $x=a$ (parallel to the $y-z$ plane) due to a point charge $q$ fixed at the origin is given as $\frac{q}{n\epsilon_0}$.

1. Understanding Electric Flux and Gauss's Law: According to Gauss's Law, the total electric flux through any closed surface enclosing a charge $q$ is given by $\Phi_{total} = \frac{q}{\epsilon_0}$. This flux emanates radially outwards from the point charge in all directions.

2. Solid Angle Concept: The total solid angle subtended by all directions around a point charge is $4\pi$ steradians. The electric flux through any surface is proportional to the solid angle subtended by that surface at the charge's location. $\Phi = \frac{q}{4\pi\epsilon_0} \Omega$, where $\Omega$ is the solid angle.

3. Analyzing the Given Surface: The surface is an infinite plane located at $x=a$ and parallel to the $y-z$ plane. The point charge $q$ is at the origin $(0,0,0)$. This infinite plane divides the entire space into two half-spaces: one where $x < a$ and another where $x > a$. Assuming $a > 0$ (as suggested by the diagram), the charge is located in the half-space $x < a$. All electric field lines originating from the charge and directed towards the positive x-axis (i.e., into the half-space $x>0$) will pass through this infinite plane at $x=a$.

4. Solid Angle Subtended by an Infinite Plane: An infinite plane, when viewed from a point not on the plane, subtends a solid angle of $2\pi$ steradians. This is because the plane essentially "cuts" space into two equal halves. The field lines going into one of these halves will pass through the plane. In this case, the plane at $x=a$ intercepts all the electric field lines that are directed into the $x>0$ half-space. The solid angle corresponding to this half-space (or this infinite plane) from the origin is $2\pi$ steradians.

5. Calculating the Electric Flux: Since the solid angle subtended by the infinite plane at the origin is $\Omega = 2\pi$, the electric flux passing through it is: $\Phi = \frac{q}{4\pi\epsilon_0} \Omega = \frac{q}{4\pi\epsilon_0} (2\pi)$ $\Phi = \frac{q}{2\epsilon_0}$

6. Comparing with the Given Flux: The problem states that the electric flux is $\frac{q}{n\epsilon_0}$. Comparing our calculated flux with the given expression: $\frac{q}{2\epsilon_0} = \frac{q}{n\epsilon_0}$ From this, we can clearly see that $n=2$.

Alternative Method (Direct Integration): The electric field due to a point charge $q$ at the origin at a point $(x,y,z)$ is given by: $\vec{E} = \frac{q}{4\pi\epsilon_0} \frac{\vec{r}}{|\vec{r}|^3} = \frac{q}{4\pi\epsilon_0} \frac{x\hat{i} + y\hat{j} + z\hat{k}}{(x^2+y^2+z^2)^{3/2}}$ For the surface at $x=a$, the differential area vector is $d\vec{A} = dy dz \hat{i}$. The electric flux $\Phi$ is given by $\iint \vec{E} \cdot d\vec{A}$: $\Phi = \iint_{y=-\infty}^{\infty} \int_{z=-\infty}^{\infty} \left( \frac{q}{4\pi\epsilon_0} \frac{a\hat{i} + y\hat{j} + z\hat{k}}{(a^2+y^2+z^2)^{3/2}} \right) \cdot (dy dz \hat{i})$ $\Phi = \frac{qa}{4\pi\epsilon_0} \iint_{y=-\infty}^{\infty} \int_{z=-\infty}^{\infty} \frac{dy dz}{(a^2+y^2+z^2)^{3/2}}$ To evaluate this integral, convert to polar coordinates in the $y-z$ plane: let $y = R\cos\theta$, $z = R\sin\theta$, so $y^2+z^2 = R^2$ and $dy dz = R dR d\theta$. The limits are $R$ from $0$ to $\infty$ and $\theta$ from $0$ to $2\pi$. $\Phi = \frac{qa}{4\pi\epsilon_0} \int_{0}^{2\pi} d\theta \int_{0}^{\infty} \frac{R dR}{(a^2+R^2)^{3/2}}$ The integral with respect to $\theta$ is $2\pi$. For the integral with respect to $R$: let $u = a^2+R^2$, so $du = 2R dR$. $\int_{0}^{\infty} \frac{R dR}{(a^2+R^2)^{3/2}} = \int_{a^2}^{\infty} \frac{1}{2} u^{-3/2} du = \frac{1}{2} \left[ \frac{u^{-1/2}}{-1/2} \right]_{a^2}^{\infty} = \left[ -u^{-1/2} \right]_{a^2}^{\infty} = \left[ -\frac{1}{\sqrt{u}} \right]_{a^2}^{\infty}$ $= -\left( \lim_{u\to\infty} \frac{1}{\sqrt{u}} - \frac{1}{\sqrt{a^2}} \right) = -\left( 0 - \frac{1}{a} \right) = \frac{1}{a}$ Substitute these results back into the flux equation: $\Phi = \frac{qa}{4\pi\epsilon_0} (2\pi) \left( \frac{1}{a} \right) = \frac{q}{2\epsilon_0}$ Comparing this with $\frac{q}{n\epsilon_0}$, we find $n=2$.

The final answer is $\boxed{2}$.

Explanation of the solution: The total electric flux from a point charge $q$ is $q/\epsilon_0$. An infinite plane passing through space effectively divides the space into two halves. Since the charge is located on one side of the plane, and the plane extends infinitely, it intercepts exactly half of the total electric field lines emanating from the charge. Therefore, the electric flux passing through the infinite plane is half of the total flux, i.e., $\frac{1}{2} \frac{q}{\epsilon_0}$. Comparing this with the given flux $\frac{q}{n\epsilon_0}$, we find $n=2$.