The electric field in a region is given by \( \vec{E} = \displaystyle\frac{E_0\,x}{b}\,\hat{\imath} \). Find the charge contained in the cubical volume bounded by the x = 0, x = a, y = 0, y = a and z = 0, z = a. Take \( E_0 = 6 \times 10^{3}\,\mathrm{N/C} \), a = 1 cm and b = 2 cm
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 2.65 \times 10^{-12}\,\mathrm{C} \)
By Gauss's law the charge inside the cube is \( q = \varepsilon_0 \Phi \), where \( \Phi \) is the net flux through its surface.
The field is along \( \hat{\imath} \), so the four faces parallel to the x-axis (\( y = 0, a \) and \( z = 0, a \)) carry no flux. On the face \( x = 0 \) the field is zero. Only the face \( x = a \) contributes, where \( E = E_0 a/b \) and the area is \( a^{2} \).
\( \displaystyle \Phi = \frac{E_0 a}{b} \times a^{2} = \frac{E_0 a^{3}}{b} \)
\( a = 1\,\mathrm{cm} = 10^{-2}\,\mathrm{m} \)
\( b = 2\,\mathrm{cm} = 2\times10^{-2}\,\mathrm{m} \)
\( \displaystyle \Phi = \frac{6\times10^{3} \times (10^{-2})^{3}}{2\times10^{-2}} = 0.3\,\mathrm{N\,m^{2}\,C^{-1}} \)
\( \displaystyle q = \varepsilon_0\Phi = 8.85\times10^{-12} \times 0.3 = 2.65\times10^{-12}\,\mathrm{C} \)