Question

There is a point charge at distance d from disc of radius .what is flux on it

Answer 1

\frac{q}{2\epsilon_0}(1 - \frac{d}{\sqrt{R^2+d^2}})

Answer 2

Let the point charge be $q$ and the disc have radius $R$. Assume the point charge is located on the axis of the disc at a distance $d$ from its center. Let the center of the disc be at the origin $(0,0,0)$ and the disc lie in the $xy$-plane. The point charge is at $(0,0,d)$.

The electric field at a point $P$ on the disc with polar coordinates $(r', \theta)$ (Cartesian coordinates $(r'\cos\theta, r'\sin\theta, 0)$) is given by $\vec{E} = \frac{q}{4\pi\epsilon_0} \frac{\vec{r}}{|\vec{r}|^3}$, where $\vec{r}$ is the vector from the charge to the point $P$. $\vec{r} = (r'\cos\theta \hat{i} + r'\sin\theta \hat{j} + 0 \hat{k}) - (0 \hat{i} + 0 \hat{j} + d \hat{k}) = r'\cos\theta \hat{i} + r'\sin\theta \hat{j} - d \hat{k}$. $|\vec{r}|^2 = (r'\cos\theta)^2 + (r'\sin\theta)^2 + (-d)^2 = (r')^2 + d^2$. $|\vec{r}| = \sqrt{(r')^2 + d^2}$.

The electric field at point $P$ is $\vec{E} = \frac{q}{4\pi\epsilon_0} \frac{r'\cos\theta \hat{i} + r'\sin\theta \hat{j} - d \hat{k}}{((r')^2 + d^2)^{3/2}}$.

To calculate the flux, we need to integrate $\vec{E} \cdot d\vec{A}$ over the surface of the disc. The differential area vector $d\vec{A}$ for a point on the disc is $dA \hat{n}$, where $\hat{n}$ is the unit vector normal to the disc. Let's choose the normal to be in the positive z-direction, so $\hat{n} = \hat{k}$. The area element in polar coordinates is $dA = r' dr' d\theta$. So, $d\vec{A} = r' dr' d\theta \hat{k}$.

The dot product $\vec{E} \cdot d\vec{A}$ is: $\vec{E} \cdot d\vec{A} = \left( \frac{q}{4\pi\epsilon_0} \frac{r'\cos\theta \hat{i} + r'\sin\theta \hat{j} - d \hat{k}}{((r')^2 + d^2)^{3/2}} \right) \cdot (r' dr' d\theta \hat{k})$ $\vec{E} \cdot d\vec{A} = \frac{q}{4\pi\epsilon_0} \frac{(-d)}{((r')^2 + d^2)^{3/2}} (r' dr' d\theta)$.

The total flux $\Phi$ through the disc is the integral of this expression over the disc surface. The radial coordinate $r'$ goes from $0$ to $R$, and the angular coordinate $\theta$ goes from $0$ to $2\pi$. $\Phi = \int_{S} \vec{E} \cdot d\vec{A} = \int_0^{2\pi} \int_0^R \frac{q}{4\pi\epsilon_0} \frac{-d r' dr' d\theta}{((r')^2 + d^2)^{3/2}}$.

We can separate the integrals: $\Phi = \frac{q}{4\pi\epsilon_0} \int_0^{2\pi} d\theta \int_0^R \frac{-d r'}{((r')^2 + d^2)^{3/2}} dr'$.

The integral with respect to $\theta$ is $\int_0^{2\pi} d\theta = 2\pi$.

The integral with respect to $r'$ is $\int_0^R \frac{-d r'}{((r')^2 + d^2)^{3/2}} dr'$. Let $u = (r')^2 + d^2$. Then $du = 2r' dr'$, so $r' dr' = \frac{1}{2} du$. When $r' = 0$, $u = 0^2 + d^2 = d^2$. When $r' = R$, $u = R^2 + d^2$. The integral becomes $\int_{d^2}^{R^2+d^2} \frac{-d}{u^{3/2}} \frac{1}{2} du = -\frac{d}{2} \int_{d^2}^{R^2+d^2} u^{-3/2} du$. $\int u^{-3/2} du = \frac{u^{-3/2 + 1}}{-3/2 + 1} = \frac{u^{-1/2}}{-1/2} = -2 u^{-1/2} = \frac{-2}{\sqrt{u}}$. So the definite integral is $-\frac{d}{2} \left[ \frac{-2}{\sqrt{u}} \right]_{d^2}^{R^2+d^2} = -d \left[ \frac{-1}{\sqrt{u}} \right]_{d^2}^{R^2+d^2} = d \left[ \frac{1}{\sqrt{u}} \right]_{d^2}^{R^2+d^2}$. $= d \left( \frac{1}{\sqrt{R^2 + d^2}} - \frac{1}{\sqrt{d^2}} \right)$. Assuming $d > 0$, $\sqrt{d^2} = d$. $= d \left( \frac{1}{\sqrt{R^2 + d^2}} - \frac{1}{d} \right) = \frac{d}{\sqrt{R^2 + d^2}} - 1$.

Now substitute this back into the expression for $\Phi$: $\Phi = \frac{q}{4\pi\epsilon_0} (2\pi) \left( \frac{d}{\sqrt{R^2 + d^2}} - 1 \right)$ $\Phi = \frac{q}{2\epsilon_0} \left( \frac{d}{\sqrt{R^2 + d^2}} - 1 \right)$.

This result is negative, which means the flux is inwards through the disc if the normal is taken in the positive z-direction (away from the charge). The question asks for "flux on it", which usually refers to the magnitude or the net outward flux. If we consider the flux as positive when field lines pass through the disc away from the charge, and negative when they pass towards the charge, then for a positive charge at $z=d$ and the disc at $z=0$, the field lines pass towards the disc, so the flux is negative. The magnitude of the flux is $|\Phi| = \frac{q}{2\epsilon_0} \left( 1 - \frac{d}{\sqrt{R^2 + d^2}} \right)$.

Alternatively, we can use the concept of the solid angle subtended by the disc at the point charge. The flux through any surface subtending a solid angle $\Omega$ at a point charge $q$ is given by $\Phi = \frac{q \Omega}{4\pi\epsilon_0}$. The solid angle $\Omega$ subtended by a disc of radius $R$ at a point on its axis at distance $d$ is given by $\Omega = 2\pi (1 - \cos\alpha)$, where $\alpha$ is the half-angle of the cone formed by the charge and the edge of the disc. From the geometry, $\cos\alpha = \frac{d}{\sqrt{R^2 + d^2}}$. So, the solid angle is $\Omega = 2\pi \left( 1 - \frac{d}{\sqrt{R^2 + d^2}} \right)$. The flux is $\Phi = \frac{q}{4\pi\epsilon_0} \Omega = \frac{q}{4\pi\epsilon_0} \left[ 2\pi \left( 1 - \frac{d}{\sqrt{R^2 + d^2}} \right) \right]$. $\Phi = \frac{q}{2\epsilon_0} \left( 1 - \frac{d}{\sqrt{R^2 + d^2}} \right)$.

This result is positive, implying the flux is considered positive when the field lines go from the charge through the surface. This corresponds to taking the normal vector pointing towards the charge if the charge is in front of the surface, or outwards from the charge if the surface encloses the charge (which is not the case here). If the charge is at $z=d$ and the disc is at $z=0$, the field lines go from $z=d$ towards $z=0$. If we take the normal of the disc pointing upwards (positive z), the dot product is negative. If we take the normal pointing downwards (negative z), the dot product is positive. The solid angle method inherently calculates the flux leaving the charge and passing through the surface, which corresponds to the choice of the normal pointing towards the charge (or away from the charge, depending on convention). The result using the solid angle is usually taken as the magnitude of the flux or the flux in the direction away from the source, which is usually what is implied.

Given the ambiguity of the question (direction of normal), both $\frac{q}{2\epsilon_0} \left( 1 - \frac{d}{\sqrt{R^2 + d^2}} \right)$ and $\frac{q}{2\epsilon_0} \left( \frac{d}{\sqrt{R^2 + d^2}} - 1 \right)$ are possible answers, depending on the chosen direction of the normal vector. However, the solid angle method gives the magnitude of the flux passing through the surface, or the flux in the direction away from the source, which is usually what is implied.

Final Answer based on standard solid angle approach: $\Phi = \frac{q}{2\epsilon_0} \left( 1 - \frac{d}{\sqrt{R^2 + d^2}} \right)$.

If the charge is at distance d from the disc, and it is implied that the charge is on the axis of the disc.