KCET · Physics · Electrostatics
A uniform electric field \(E=3 \times 10^5 \mathrm{NC}^{-1}\) is acting along the positive \(Y\)-axis. The electric flux through a rectangle of area \(10 \mathrm{~cm} \times 30 \mathrm{~cm}\) whose plane is parallel to the ZX -plane is
- A \(12 \times 10^3 \mathrm{Vm}\)
- B \(9 \times 10^3 \mathrm{Vm}\)
- C \(15 \times 10^3 \mathrm{Vm}\)
- D \(18 \times 10^5 \mathrm{Vm}\)
Answer & Solution
Correct Answer
(B) \(9 \times 10^3 \mathrm{Vm}\)
Step-by-step Solution
Detailed explanation
Given, \(\mathbf{E}=3 \times 10^5 \mathbf{\mathbf { j }} \mathrm{NC}^{-1}\)
\(A=10 \mathrm{~cm} \times 30 \mathrm{~cm}=3 \times 10^{-2} \mathrm{~m}^2\)
\(\therefore \quad \mathbf{A}=3 \times 10^{-2} \hat{\mathbf{j}} \mathrm{~m}^2\)
(For ZX-plane)
\(\therefore\) Electric flux, \(\phi=\mathbf{E} \cdot \mathbf{A}\)
\(=3 \times 10^5 \mathbf{j} \times 3 \times 10^{-2} \hat{\mathbf{j}}\)
\(=9 \times 10^3 \mathrm{~V}-\mathrm{m}\)
\(A=10 \mathrm{~cm} \times 30 \mathrm{~cm}=3 \times 10^{-2} \mathrm{~m}^2\)
\(\therefore \quad \mathbf{A}=3 \times 10^{-2} \hat{\mathbf{j}} \mathrm{~m}^2\)
(For ZX-plane)
\(\therefore\) Electric flux, \(\phi=\mathbf{E} \cdot \mathbf{A}\)
\(=3 \times 10^5 \mathbf{j} \times 3 \times 10^{-2} \hat{\mathbf{j}}\)
\(=9 \times 10^3 \mathrm{~V}-\mathrm{m}\)
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