Q160 / 180 · KCET 2014 · Physics
What is the electric potential at a distance of \( 9 \mathrm{~cm} \) from \( 3 \mathrm{nC} ? \)
- A\( 270 \mathrm{~V} \)
- B\( 3 \mathrm{~V} \)
- Correct answerC\( 300 \mathrm{~V} \)
- D\( 30 \mathrm{~V} \)
Solution
Electric potential is given as
\[ V=\frac{1}{4 \Pi \varepsilon_{0}} \frac{q}{r} \]
Given
\[ q=3 n C=3 \times 10^{-9} C ; r=9 \mathrm{~cm}=9 \times 10^{-2} \mathrm{~m} \]
Therefore,
\[ V=\left(9 \times 10^{\circ}\right) \frac{3 \times 10^{-9}}{9 \times 10^{-2}}=3 \times 10^{2} V \]
Thus, electric potential at a distance of \( 9 \mathrm{~cm} \) from \( 3 \mathrm{nC} \) is \( 300 \mathrm{~V} \)
\[ V=\frac{1}{4 \Pi \varepsilon_{0}} \frac{q}{r} \]
Given
\[ q=3 n C=3 \times 10^{-9} C ; r=9 \mathrm{~cm}=9 \times 10^{-2} \mathrm{~m} \]
Therefore,
\[ V=\left(9 \times 10^{\circ}\right) \frac{3 \times 10^{-9}}{9 \times 10^{-2}}=3 \times 10^{2} V \]
Thus, electric potential at a distance of \( 9 \mathrm{~cm} \) from \( 3 \mathrm{nC} \) is \( 300 \mathrm{~V} \)
Hence, the correct answer is option: 'C'.