KCET · Physics · Communication System
\( 4 \times 10^{10} \) electrons are removed from a neutral metal sphere of diameter \( 20 \mathrm{~cm} \) placed in air.
The magnitude of the electric field (in \( N C^{-1} \) ) at a distance of \( 20 \mathrm{~cm} \) from its centre is:
- A \( 5760 \)
- B \( 1440 \)
- C \( 640 \)
- D Zero
Answer & Solution
Correct Answer
(B) \( 1440 \)
Step-by-step Solution
Detailed explanation
Given, number of electrons removed \(=4 \times 10^{10}\); diameter of metal sphere \(=20 \mathrm{~cm}\); distance \(\mathrm{r}=20 \mathrm{~cm}\). Therefore,
Electric field \(=\frac{1}{4 \Pi \varepsilon_{0}} \frac{q}{r^{2}}\)
where \(\frac{1}{4 \Pi \varepsilon_{0}}=9 \times 10^{9}\)
\(\Rightarrow q=\left(4 \times 10^{10}\right) \times\left(1.6 \times 10^{-19}\right)=6.4 \times 10^{-9}\)
Electric field
\(=\frac{9 \times 10^{9} \times 6.4 \times 10^{-9}}{\left(20 \times 10^{-2}\right)^{2}}=1440 \mathrm{~N} \mathrm{C}^{-1}\)
Electric field \(=\frac{1}{4 \Pi \varepsilon_{0}} \frac{q}{r^{2}}\)
where \(\frac{1}{4 \Pi \varepsilon_{0}}=9 \times 10^{9}\)
\(\Rightarrow q=\left(4 \times 10^{10}\right) \times\left(1.6 \times 10^{-19}\right)=6.4 \times 10^{-9}\)
Electric field
\(=\frac{9 \times 10^{9} \times 6.4 \times 10^{-9}}{\left(20 \times 10^{-2}\right)^{2}}=1440 \mathrm{~N} \mathrm{C}^{-1}\)
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