MHT CET · Physics · Electrostatics
A uniformly charged conducting sphere of diameter 14 cm has surface charge density of \(40 \mu \mathrm{Cm}^{-2}\). The total electric flux leaving the surface of the sphere is nearly (Permittivity of free space \(=8.85 \times 10^{-12}\) SI unit)
- A 40 kWb
- B 140 kWb
- C 240 kWb
- D 280 kWb
Answer & Solution
Correct Answer
(D) 280 kWb
Step-by-step Solution
Detailed explanation
Electric flux \(\phi=\frac{\mathrm{q}}{\varepsilon_0}=\frac{4 \pi \mathrm{r}^2 \sigma}{\varepsilon_0}\)
\(\begin{aligned} & =\frac{4 \times 3.14 \times\left(7 \times 10^{-2}\right)^2 \times 40 \times 10^{-6}}{8.85 \times 10^{-12}} \\ & =\frac{4 \times 3.14 \times 49 \times 10^{-4} \times 4 \times 10^{-5}}{8.85 \times 10^{-12}} \\ & =\frac{4 \times 3.14 \times 49 \times 4 \times 10^3}{8.85} \\ & =278 \times 10^3 \approx 280 \times 10^3 \mathrm{~Wb} \\ & =280 \mathrm{kWb}\end{aligned}\)
\(\begin{aligned} & =\frac{4 \times 3.14 \times\left(7 \times 10^{-2}\right)^2 \times 40 \times 10^{-6}}{8.85 \times 10^{-12}} \\ & =\frac{4 \times 3.14 \times 49 \times 10^{-4} \times 4 \times 10^{-5}}{8.85 \times 10^{-12}} \\ & =\frac{4 \times 3.14 \times 49 \times 4 \times 10^3}{8.85} \\ & =278 \times 10^3 \approx 280 \times 10^3 \mathrm{~Wb} \\ & =280 \mathrm{kWb}\end{aligned}\)
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