Animated Solution for Physics - Electrostatics: An electric field $\mathbf{E}=4x\hat{\mathbf{i}}-(y^2+1)\hat{\mathbf{j}}$ N/C passes through the box shown in figure. The flux of the electric field through surfaces $ABCD$ and $BCGF$ are marked as ${\varphi }_I$ and ${\varphi }_{II}$, respectively. The difference between $({\varphi }_I-{\varphi }_{II})$ is (in ${\text{N-m}}^2/\text{C}$) ...... .

Imagine a rectangular box placed in a 3D coordinate system. We are given an electric field $\mathbf{E}=4x\hat{\mathbf{i}}-(y^2+1)\hat{\mathbf{j}}$ N/C passing through this box. Our goal is to find the electric flux through two specific faces: the top face $ABCD$ (denoted as ${\varphi }_I$) and the right face $BCGF$ (denoted as ${\varphi }_{II}$), and then calculate their difference.

Electric flux through any flat surface is mathematically defined as the dot product of the electric field vector and the area vector:

To solve this, we must carefully determine the area vector for each face. Remember, for a closed 3D object, the area vector always points strictly outwards, normal to the surface.

Let's focus on the top face, $ABCD$. By observing the coordinates, we can see it lies horizontally at a height of $z=2$. Its length along the X-axis is $3$ units, and its width along the Y-axis is $2$ units.

The area of this rectangle is simply $3\times 2=6$. Because it is the top face of the box, its outward normal points straight up, along the positive Z-axis. Therefore, the area vector ${\mathbf{A}}_1$ is:

Now, we calculate the flux ${\varphi }_I$ by taking the dot product of the electric field and ${\mathbf{A}}_1$:

Notice that the electric field only has $\hat{\mathbf{i}}$ and $\hat{\mathbf{j}}$ components. There is no $\hat{\mathbf{k}}$ component! Because the dot product of perpendicular unit vectors is zero ($\hat{\mathbf{i}}\cdot \hat{\mathbf{k}}=0$ and $\hat{\mathbf{j}}\cdot \hat{\mathbf{k}}=0$), the flux through the top face is simply:

Next, let's examine the right face, $BCGF$. This face lies in the vertical plane where $x=3$. Its dimensions are $2$ units along the Y-axis and $2$ units along the Z-axis.

The area is $2\times 2=4$. The outward normal for this right face points along the positive X-axis. So, the area vector ${\mathbf{A}}_2$ is:

Let's calculate the flux ${\varphi }_{II}$:

This time, the $\hat{\mathbf{i}}$ component of the electric field multiplies with the $4\hat{\mathbf{i}}$ of the area vector, giving us $16x$.

Here is the crucial catch: This entire face is located at the specific coordinate $x=3$. We must substitute this value into our flux expression:

Finally, the question asks for the difference between ${\varphi }_I$ and ${\varphi }_{II}$.

And that is our final, elegant answer!