Question:
A cube is placed inside an electric field, \(\vec{E}=150y^2\hat{j}\). The side of the cube is 0.5 m and is placed in the field as shown in the given figure. The charge inside the cube is :
A cube is placed inside an electric field, \(\vec{E}=150y^2\hat{j}\). The side of the cube is 0.5 m and is placed in the field as shown in the given figure. The charge inside the cube is :
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The field is along y and depends only on y, so flux passes only through the top (y = 0.5 m) and bottom (y = 0) faces; then \(q=\varepsilon_0\phi\).
Updated On: Nov 4, 2026
- \(3.8\times10^{-11}\ \mathrm{C}\)
- \(8.3\times10^{-11}\ \mathrm{C}\)
- \(3.8\times10^{-12}\ \mathrm{C}\)
- \(8.3\times10^{-12}\ \mathrm{C}\)
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The Correct Option is B
Solution and Explanation
Step 1: The field \(\vec E=150y^2\hat j\) is parallel to the y-axis, so there is no flux through the four side faces. The bottom face of the cube lies at \(y=0\) and the top face at \(y=0.5\) m (as in the figure).
Step 2: Face area \(A=(0.5)^2=0.25\ \mathrm{m^2}\).
Step 3: Flux through top (outward normal \(+\hat j\)): \(E_{top}A=150(0.5)^2(0.25)=9.375\). Flux through bottom: \(-E(0)A=0\).
Step 4: Net flux \(\phi=9.375\ \mathrm{N\,m^2/C}\).
Step 5: Gauss's law: \(q=\varepsilon_0\phi=8.85\times10^{-12}\times9.375\approx8.3\times10^{-11}\ \mathrm{C}\).
Answer: (2) \(8.3\times10^{-11}\ \mathrm{C}\)
Step 2: Face area \(A=(0.5)^2=0.25\ \mathrm{m^2}\).
Step 3: Flux through top (outward normal \(+\hat j\)): \(E_{top}A=150(0.5)^2(0.25)=9.375\). Flux through bottom: \(-E(0)A=0\).
Step 4: Net flux \(\phi=9.375\ \mathrm{N\,m^2/C}\).
Step 5: Gauss's law: \(q=\varepsilon_0\phi=8.85\times10^{-12}\times9.375\approx8.3\times10^{-11}\ \mathrm{C}\).
Answer: (2) \(8.3\times10^{-11}\ \mathrm{C}\)
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