Question 712: Electrostats & Capacitance

Question

An electric field, $$\overrightarrow{\mathrm{E}}=\frac{2 \hat{i}+6 \hat{j}+8 \hat{k}}{\sqrt{6}}$$ passes through the surface of $$4 \mathrm{~m}^2$$ area having unit vector $$\hat{n}=\left(\frac{2 \hat{i}+\hat{j}+\hat{k}}{\sqrt{6}}\right)$$. The electric flux for that surface is _________ $$\mathrm{Vm}$$.

Accepted Answer

The electric flux through a surface is given by the formula:
$$\Phi = \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{A}} = |\overrightarrow{\mathrm{E}}||\overrightarrow{\mathrm{A}}|\cos	heta$$
where $$\overrightarrow{\mathrm{E}}$$ is the electric field, $$\overrightarrow{\mathrm{A}}$$ is the area vector (with magnitude equal to the area of the surface and direction perpendicular to the surface, defined by the unit vector $$\hat{n}$$), and $$	heta$$ is the angle between $$\overrightarrow{\mathrm{E}}$$ and $$\overrightarrow{\mathrm{A}}$$. However, when using unit vectors to describe the directions of $$\overrightarrow{\mathrm{E}}$$ and $$\hat{n}$$, the dot product can be used to simplify the calculation as follows:
$$\Phi = \overrightarrow{\mathrm{E}} \cdot \left(\overrightarrow{\mathrm{A}}ight) = (\overrightarrow{\mathrm{E}} \cdot \hat{n})A$$
Given that $$\overrightarrow{\mathrm{E}}=\frac{2 \hat{i}+6 \hat{j}+8 \hat{k}}{\sqrt{6}}$$ and $$\hat{n}=\left(\frac{2 \hat{i}+\hat{j}+\hat{k}}{\sqrt{6}}ight)$$, and the area $$A = 4 \mathrm{m}^2$$, we can substitute them into our formula. Note that since $$\overrightarrow{\mathrm{A}} = A\hat{n}$$, the magnitude of the area vector is the area of the surface itself. First, let's find $$\overrightarrow{\mathrm{E}} \cdot \hat{n}$$:
$$\overrightarrow{\mathrm{E}} \cdot \hat{n} = \left(\frac{2 \hat{i}+6 \hat{j}+8 \hat{k}}{\sqrt{6}}ight) \cdot \left(\frac{2 \hat{i}+\hat{j}+\hat{k}}{\sqrt{6}}ight)$$
To compute the dot product, we multiply corresponding components and then add them up:
$$\left(\frac{2 \hat{i}+6 \hat{j}+8 \hat{k}}{\sqrt{6}}ight) \cdot \left(\frac{2 \hat{i}+\hat{j}+\hat{k}}{\sqrt{6}}ight) = \frac{1}{6}(2\cdot2 + 6\cdot1 + 8\cdot1)$$
$$= \frac{1}{6}(4 + 6 + 8) = \frac{18}{6} = 3$$
Then, the electric flux through the surface is:
$$\Phi = (\overrightarrow{\mathrm{E}} \cdot \hat{n})A = 3 	imes 4 \mathrm{m}^2$$
$$\Phi = 12 \mathrm{Vm}$$
So, the electric flux for that surface is $$12 \mathrm{Vm}$$.

Electrostats & Capacitance

An electric field, $\overrightarrow{\mathrm{E}}=\frac{2 \hat{i}+6 \hat{j}+8 \hat{k}}{\sqrt{6}} passes through the surface of 4 \mathrm{~m}^2 area having unit vector \hat{n}=\left(\frac{2 \hat{i}+\hat{j}+\hat{k}}{\sqrt{6}}\right). The electric flux for that surface is _________ \mathrm{Vm}$.

Choose the correct answer: