KCET · Physics · Electrostatics
A cubical Gaussian surface has side of length \(a=10 \mathrm{~cm}\). Electric field lines are parallel to \(X\)-axis as shown in figure. The magnitudes of electric fields through surfaces \(A B C D\) and \(E F G H\) are \(6 \mathrm{kNC}^{-1}\) and \(9 \mathrm{kNC}^{-1}\) respectively. Then, the total charge enclosed by the cube is
[Take, \(\varepsilon_0=9 \times 10^{-12} \mathrm{Fm}^{-1}\) ]
- A \(-0.27 \mathrm{nC}\)
- B \(1.35 \mathrm{nC}\)
- C \(-1.35 \mathrm{nC}\)
- D \(0.27 \mathrm{nC}\)
Answer & Solution
Correct Answer
(D) \(0.27 \mathrm{nC}\)
Step-by-step Solution
Detailed explanation
Total flux \(\phi_E=E \cdot d s\)
\(=9 \mathrm{k}-6 \mathrm{k} \times\left(10^{-2} \times 10\right)^2\)
\(=3 \times 10^3 \times 10^{-2}=30 \mathrm{NC}^{-1} \mathrm{~m}^2\)
According to Gauss' law, \(\frac{q}{\varepsilon_0}=\phi_E\)
\(\Rightarrow q_{\text {endosed }} =3 \times \varepsilon_0 \)
\( =30 \times 9 \times 10^{-12} \mathrm{C}=270 \times 10^{-9} \times 10^{-3} \)
\( =0.27 \times 10^{-9} \mathrm{C}=0.27 \mathrm{nC}\)
\(=9 \mathrm{k}-6 \mathrm{k} \times\left(10^{-2} \times 10\right)^2\)
\(=3 \times 10^3 \times 10^{-2}=30 \mathrm{NC}^{-1} \mathrm{~m}^2\)
According to Gauss' law, \(\frac{q}{\varepsilon_0}=\phi_E\)
\(\Rightarrow q_{\text {endosed }} =3 \times \varepsilon_0 \)
\( =30 \times 9 \times 10^{-12} \mathrm{C}=270 \times 10^{-9} \times 10^{-3} \)
\( =0.27 \times 10^{-9} \mathrm{C}=0.27 \mathrm{nC}\)
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