Question

Electric field in a region is found to be $\vec{E} = 3y\hat{j}$

The total energy stored in electric field inside the cube shown will be :

• A. $9a^5\epsilon_0$
• B. $3\epsilon_0a^3$
• C. $\frac{3}{2}\epsilon_0a^5$
• D. 0

Answer 1

Answer: (C)

$\frac{3}{2}\epsilon_0 a^5$

Answer 2

The total energy stored in an electric field in a given volume V is given by the integral of the energy density over that volume:

$U_E = \int_V u_E dV$

where $u_E$ is the electric energy density, defined as:

$u_E = \frac{1}{2}\epsilon_0 E^2$

Given electric field is $\vec{E} = 3y\hat{j}$. The magnitude of the electric field is $E = |\vec{E}| = |3y| = 3y$ (since y is positive in the cube from 0 to a). So, $E^2 = (3y)^2 = 9y^2$.

The cube shown in the figure has side length 'a'. Assuming the origin O is at one corner, the cube extends from $x=0$ to $x=a$, $y=0$ to $y=a$, and $z=0$ to $z=a$. The volume element in Cartesian coordinates is $dV = dx dy dz$.

Now, we can set up the integral for the total energy: $U_E = \int_0^a \int_0^a \int_0^a \frac{1}{2}\epsilon_0 (9y^2) dx dy dz$

We can pull out the constants from the integral: $U_E = \frac{9}{2}\epsilon_0 \int_0^a \int_0^a \int_0^a y^2 dx dy dz$

The integral can be separated into three independent integrals: $U_E = \frac{9}{2}\epsilon_0 \left( \int_0^a dx \right) \left( \int_0^a y^2 dy \right) \left( \int_0^a dz \right)$

Evaluate each integral:

1. $\int_0^a dx = [x]_0^a = a - 0 = a$

2. $\int_0^a y^2 dy = \left[ \frac{y^3}{3} \right]_0^a = \frac{a^3}{3} - \frac{0^3}{3} = \frac{a^3}{3}$

3. $\int_0^a dz = [z]_0^a = a - 0 = a$

Substitute these results back into the expression for $U_E$: $U_E = \frac{9}{2}\epsilon_0 (a) \left( \frac{a^3}{3} \right) (a)$ $U_E = \frac{9}{2}\epsilon_0 \frac{a^5}{3}$ $U_E = \frac{3}{2}\epsilon_0 a^5$