KCET · Physics · Current Electricity
What is the electric potential at a distance of \( 9 \mathrm{~cm} \) from \( 3 \mathrm{nC} ? \)
- A \( 270 \mathrm{~V} \)
- B \( 3 \mathrm{~V} \)
- C \( 300 \mathrm{~V} \)
- D \( 30 \mathrm{~V} \)
Answer & Solution
Correct Answer
(C) \( 300 \mathrm{~V} \)
Step-by-step Solution
Detailed explanation
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 \( 300 \mathrm{~V} \)
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