If 5.0 μC charge is placed at the center of a Gaussian surface in
A. 5.6 × 10<sup>5</sup> N m<sup>2</sup>/C
B. 1.4 × 10<span style="position: relative; line-height: 0; vertical-align: baseline; top: -0.5em;font-size:10.5px;">5</span> N m<span style="position: relative; line-height: 0; vertical-align: baseline; top: -0.5em;font-size:10.5px;">2</span>/C
C. 9.4 × 10<sup>4</sup> N m<span style="position: relative; line-height: 0; vertical-align: baseline; top: -0.5em;font-size:10.5px;">2</span>/C
D. 1.9 × 10<span style="position: relative; line-height: 0; vertical-align: baseline; top: -0.5em;font-size:10.5px;">4</span> N m<span style="position: relative; line-height: 0; vertical-align: baseline; top: -0.5em; font-size:10.5px;">2</span>/C
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Right Answer is: C
SOLUTION
Gauss’s Law:
The total of the electric flux out of a closed surface is equal to the charge (Q) enclosed divided by the permittivity (ϵ0) of the medium. It can be given by ϕE
\({ϕ _{\bf{E}}} = \frac{{\bf{Q}}}{{{\epsilon_0}}}\)
Analysis:
A charge is placed at the center of the cube as shown:
Flux is linked equally with all the faces (6 faces)
∴ Using Gauss’s theorem, the flux linked with 1 face will be:
\({{ϕ }_{E}}=\frac{q}{{{\epsilon }_{0}}}\)
As a cube is given, flux linked with 6 faces are:
\(6× {{ϕ }_{E}}=\frac{q}{{{\epsilon }_{0}}}\)
\({{ϕ }_{E}}=\frac{q}{6~{{\epsilon }_{0}}}\)
Calculation:
With q = 5.0 μC, the electric flux through any one face of the cube will be:
\({{ϕ }_{E}}=\frac{5× 10^{-6}}{6~{× 8.854× 10^{-12}}}\)
ϕE = 9.4 × 104 N m2/C
