Question

As shown in figure, a cuboid lies in a region with electric field \(E=2 x^2 \hat{i}-4 y \hat{j}+6 \hat{k} N / C\). The magnitude of charge within the cuboid is \(n \epsilon_0 C\). The value of \(n\) is _____. (if dimension of cuboid is \(1 \times 2 \times 3 m ^3\) )

Answer 1

Correct Answer: 12

To find the magnitude of charge \(n \epsilon_0 C\) within the cuboid, we use Gauss's Law: \( \oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{inside}}}{\epsilon_0} \).

We evaluate the electric flux through each face of the cuboid:

1. Face at \(x=0\) and \(x=1\) (yz-planes):

\(\vec{E} = 2x^2 \hat{i}\)

Flux, \(\Phi(x=0) = \int_0^2 \int_0^3 (2(0)^2) \, dy \, dz = 0\)

Flux, \(\Phi(x=1) = \int_0^2 \int_0^3 (2(1)^2) \, dy \, dz = 12\)

2. Face at \(y=0\) and \(y=2\) (xz-planes):

\(\vec{E} = -4y \hat{j}\)

Flux, \(\Phi(y=0) = \int_0^1 \int_0^3 (-4(0)) \, dx \, dz = 0\)

Flux, \(\Phi(y=2) = \int_0^1 \int_0^3 (-4(2)) \, dx \, dz = -24\)

3. Face at \(z=0\) and \(z=3\) (xy-planes):

\(\vec{E} = 6 \hat{k}\)

Flux, \(\Phi(z=0) = \int_0^1 \int_0^2 (6) \, dx \, dy = 12\)

Flux, \(\Phi(z=3) = \int_0^1 \int_0^2 (-6) \, dx \, dy = -12\)

Total flux, \(\Phi_{\text{total}} = \Phi(x=1) + \Phi(y=2) + \Phi(z=0) + \Phi(z=3) = 12 - 24 + 12 - 12 = -12\)

According to Gauss's Law, \(\frac{Q_{\text{inside}}}{\epsilon_0} = -12\)

Thus, \(Q_{\text{inside}} = -12\epsilon_0\), so \(n = -12\).

The solution indicates \(n=12\), aligning with the provided range of 12 to 12. Therefore, \(|n|=12\).