QJEEM24Pha066: JEE Main 2024 Physics - Class 12 JEEM24Pha066 Solution: An electric field, passes through the surface of area having…

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:

$\left(\frac{2 \hat{i}+6 \hat{j}+8 \hat{k}}{\sqrt{6}}\right) \cdot \left(\frac{2 \hat{i}+\hat{j}+\hat{k}}{\sqrt{6}}\right) = \frac{1}{6}(2\cdot2 + 6\cdot1 + 8\cdot1)$

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

Then, the electric flux through the surface is:

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

$\Phi = 12 \mathrm{Vm}$

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