Question

Find flux through the given surface due to a charge q kept at (0, 0, 4a) as shown.

Answer 1

q/(40ε₀)

Answer 2

To find the flux through the given surface, we will use the definition of electric flux:

$$\Phi = \int \vec{E} \cdot d\vec{A}$$

The charge q is located at (0, 0, 4a). Let this be P_q = (0, 0, 4a). The surface is a quarter-circle in the xy-plane (z=0), bounded by x=0, y=0, and the circular arc x^2 + y^2 = (3a)^2. The radius of this quarter-disk is R = 3a.

1. Determine the Electric Field ($\vec{E}$): The electric field $\vec{E}$ due to a point charge q at position $\vec{r}_q$ at a point $\vec{r}$ is given by:

$$\vec{E} = \frac{1}{4\pi\varepsilon_0} \frac{q}{|\vec{r} - \vec{r}_q|^3} (\vec{r} - \vec{r}_q)$$

Let a point on the surface be $\vec{r} = (x, y, 0)$. Then, $\vec{r} - \vec{r}_q = (x - 0)\hat{i} + (y - 0)\hat{j} + (0 - 4a)\hat{k} = x\hat{i} + y\hat{j} - 4a\hat{k}$. The magnitude squared is $|\vec{r} - \vec{r}_q|^2 = x^2 + y^2 + (-4a)^2 = x^2 + y^2 + 16a^2$. So, the electric field at a point (x, y, 0) on the surface is:

$$\vec{E} = \frac{q}{4\pi\varepsilon_0} \frac{x\hat{i} + y\hat{j} - 4a\hat{k}}{(x^2 + y^2 + 16a^2)^{3/2}}$$

2. Determine the Area Vector ($d\vec{A}$): The surface lies in the z=0 plane. The differential area vector $d\vec{A}$ can be taken as $dx dy \hat{k}$ (normal pointing in the positive z-direction).

3. Calculate the Dot Product ($\vec{E} \cdot d\vec{A}$):

$$\vec{E} \cdot d\vec{A} = \left( \frac{q}{4\pi\varepsilon_0} \frac{x\hat{i} + y\hat{j} - 4a\hat{k}}{(x^2 + y^2 + 16a^2)^{3/2}} \right) \cdot (dx dy \hat{k})$$ $$\vec{E} \cdot d\vec{A} = \frac{q}{4\pi\varepsilon_0} \frac{-4a}{(x^2 + y^2 + 16a^2)^{3/2}} dx dy$$

4. Set up the Integral: The flux $\Phi$ is the integral of this expression over the quarter-disk surface. It is convenient to use cylindrical coordinates for this integration. In cylindrical coordinates:

• $x^2 + y^2 = \rho^2$
• $dx dy = \rho d\rho d\phi$

The limits of integration for the quarter-disk in the first quadrant are:

• $\rho$ from $0$ to $3a$ (radius of the disk)
• $\phi$ from $0$ to $\pi/2$ (for the first quadrant)

$$\Phi = \int_{0}^{\pi/2} \int_{0}^{3a} \frac{q}{4\pi\varepsilon_0} \frac{-4a}{(\rho^2 + 16a^2)^{3/2}} \rho d\rho d\phi$$

We can pull the constants out of the integral:

$$\Phi = \frac{-4aq}{4\pi\varepsilon_0} \int_{0}^{\pi/2} d\phi \int_{0}^{3a} \frac{\rho}{(\rho^2 + 16a^2)^{3/2}} d\rho$$

5. Evaluate the Integrals: First, evaluate the integral with respect to $\phi$:

$$\int_{0}^{\pi/2} d\phi = [\phi]_{0}^{\pi/2} = \frac{\pi}{2}$$

Next, evaluate the integral with respect to $\rho$:

$$I_{\rho} = \int_{0}^{3a} \frac{\rho}{(\rho^2 + 16a^2)^{3/2}} d\rho$$

Let $u = \rho^2 + 16a^2$. Then $du = 2\rho d\rho$, so $\rho d\rho = \frac{1}{2} du$. When $\rho = 0$, $u = 0^2 + 16a^2 = 16a^2$. When $\rho = 3a$, $u = (3a)^2 + 16a^2 = 9a^2 + 16a^2 = 25a^2$.

$$I_{\rho} = \int_{16a^2}^{25a^2} \frac{1}{2} u^{-3/2} du$$ $$I_{\rho} = \frac{1}{2} \left[ \frac{u^{-1/2}}{-1/2} \right]_{16a^2}^{25a^2} = - \left[ \frac{1}{\sqrt{u}} \right]_{16a^2}^{25a^2}$$ $$I_{\rho} = - \left( \frac{1}{\sqrt{25a^2}} - \frac{1}{\sqrt{16a^2}} \right) = - \left( \frac{1}{5a} - \frac{1}{4a} \right)$$ $$I_{\rho} = - \left( \frac{4 - 5}{20a} \right) = - \left( \frac{-1}{20a} \right) = \frac{1}{20a}$$

6. Combine the Results: Substitute the evaluated integrals back into the flux equation:

$$\Phi = \frac{-4aq}{4\pi\varepsilon_0} \left( \frac{\pi}{2} \right) \left( \frac{1}{20a} \right)$$ $$\Phi = \frac{-4aq}{4\pi\varepsilon_0} \frac{\pi}{40a}$$ $$\Phi = \frac{-q}{4\pi\varepsilon_0} \frac{4a\pi}{40a} = \frac{-q}{4\pi\varepsilon_0} \frac{\pi}{10}$$ $$\Phi = \frac{-q}{40\varepsilon_0}$$

The negative sign indicates that the electric field lines are entering the surface, given that the normal vector was chosen in the positive z-direction and the charge is at positive z. If the question asks for the magnitude of the flux, it would be $q/(40\varepsilon_0)$.

The final answer is $\frac{q}{40\varepsilon_0}$.

Explanation of the solution: The electric flux through an open surface is calculated by integrating the dot product of the electric field and the differential area vector over the surface.

1. Define Electric Field: The electric field $\vec{E}$ due to the point charge q at (0, 0, 4a) is determined at a generic point (x, y, 0) on the surface.
2. Define Area Vector: The surface is in the xy-plane, so its differential area vector is $d\vec{A} = dx dy \hat{k}$.
3. Calculate Dot Product: Compute $\vec{E} \cdot d\vec{A}$. This gives the component of the electric field perpendicular to the surface.
4. Set up Integral: Convert the Cartesian coordinates to cylindrical coordinates ($\rho, \phi$) for easier integration, as the surface is a quarter-disk. The limits for $\rho$ are 0 to 3a, and for $\phi$ are 0 to π/2.
5. Evaluate Integral: Perform the integration step by step, first with respect to $\phi$ and then with respect to $\rho$ using a substitution method.
6. Final Result: Combine the results of the integration to obtain the total electric flux. The negative sign indicates that the field lines are entering the surface if the normal is chosen as positive z. The magnitude of the flux is $q/(40\varepsilon_0)$.