Question

A cubical volume is bounded by the surfaces \( x = 0 \), \( x = a \), \( y = 0 \), \( y = a \), \( z = 0 \), \( z = a \). The electric field in the region is given by: \[ \vec{E} = E_0 x \hat{i} \]
Where \( E_0 = 4 \times 10^4 \, \text{NC}^{-1} \, \text{m}^{-1} \). If \( a = 2 \, \text{cm} \), the charge contained in the cubical volume is \( Q \times 10^{-14} \, \text{C} \). The value of \( Q \) is ______. (Take \( \epsilon_0 = 9 \times 10^{-12} \, \text{C}^2/\text{Nm}^2 \))

Hint:

Remember Gauss’s law and the formula for electric flux. Pay close attention to the direction of the electric field and the area vector when calculating the flux.

Answer 1

Correct Answer: 288

The problem involves finding the charge enclosed in a cube using Gauss's Law, which states: \(\Phi_E = \frac{Q_{\text{enclosed}}}{\epsilon_0}\). Here, \(\Phi_E\) is the electric flux through the cube, and \(Q_{\text{enclosed}}\) is the charge we need to find.

The electric field is given by: \(\vec{E} = E_0 x \hat{i}\) where \(E_0 = 4 \times 10^4 \, \text{NC}^{-1} \, \text{m}^{-1}\).

Convert \(a\) from cm to meters: \(a = 2\,\text{cm} = 0.02\,\text{m}\).

The electric flux through a face parallel to the yz-plane (at \(x=a\)) is:

\(\Phi_E = \int \vec{E} \cdot d\vec{A} = \int_{0}^{a} E_0 x \cdot a^2 \, dy \, dz = E_0 \cdot a \cdot a^2 = E_0 a^3\).

Since \(\vec{E}\) is zero at \(x=0\) face and across the other faces (where it is perpendicular), the total flux is determined solely by the face at \(x = a\):

\(\Phi_E = E_0 a^3 = 4 \times 10^4 \times (0.02)^3 \, \text{Nm}^2/\text{C} = 32 \times 10^{-6} \, \text{Nm}^2/\text{C}\).

Using Gauss's Law, we have:

\(\Phi_E = \frac{Q_{\text{enclosed}}}{\epsilon_0} \Rightarrow Q_{\text{enclosed}} = \Phi_E \cdot \epsilon_0\).

Calculating \(Q_{\text{enclosed}}\):

\(Q_{\text{enclosed}} = 32 \times 10^{-6} \cdot 9 \times 10^{-12} \, \text{C} = 288 \times 10^{-18} \, \text{C}\).

Express \(Q_{\text{enclosed}}\) in terms of \(Q\):

\(Q_{\text{enclosed}} = Q \times 10^{-14} \Rightarrow Q = \frac{288 \times 10^{-18}}{10^{-14}} = 288\).

Thus, the value of \(Q\) is 288, fitting perfectly within the provided range (288,288).