JEE Main 2024 — Electrostatics Question with Solution

The electric flux through a surface is given by the formula:

$$\Phi =\overrightarrow{\mathrm{E}}\cdot \overrightarrow{\mathrm{A}}=|\overrightarrow{\mathrm{E}}||\overrightarrow{\mathrm{A}}|\cos \theta$$

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 $$\theta$$ 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:

$$\Phi =\overrightarrow{\mathrm{E}}\cdot \left(\overrightarrow{\mathrm{A}}\right)=(\overrightarrow{\mathrm{E}}\cdot \hat{n})A$$

Given that $$\overrightarrow{\mathrm{E}}=\frac{2\hat{i}+6\hat{j}+8\hat{k}}{\sqrt{6}}$$ and $$\hat{n}=\left(\frac{2\hat{i}+\hat{j}+\hat{k}}{\sqrt{6}}\right)$$, 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 $$\overrightarrow{\mathrm{E}}\cdot \hat{n}$$:

$$\overrightarrow{\mathrm{E}}\cdot \hat{n}=\left(\frac{2\hat{i}+6\hat{j}+8\hat{k}}{\sqrt{6}}\right)\cdot \left(\frac{2\hat{i}+\hat{j}+\hat{k}}{\sqrt{6}}\right)$$

To compute the dot product, we multiply corresponding components and then add them up:

$$\left(\frac{2\hat{i}+6\hat{j}+8\hat{k}}{\sqrt{6}}\right)\cdot \left(\frac{2\hat{i}+\hat{j}+\hat{k}}{\sqrt{6}}\right)=\frac{1}{6}(2\cdot 2+6\cdot 1+8\cdot 1)$$

$$=\frac{1}{6}(4+6+8)=\frac{18}{6}=3$$

Then, the electric flux through the surface is:

$$\Phi =(\overrightarrow{\mathrm{E}}\cdot \hat{n})A=3\times 4{\mathrm{m}}^2$$

$$\Phi =12\mathrm{Vm}$$

So, the electric flux for that surface is $$12\mathrm{Vm}$$.