Question

The electric potential is +100 V at a distance of 10 cm from a point charge \(q\). Then, \(q\) is equal to

• A. \(+1.1 \times 10^{-9}\) C
• B. \(+1.1 \times 10^{-3}\) C
• C. 3 C
• D. \(3 \times 10^{-9}\) C

Hint:

\(V = \dfrac{kq}{r}\) where \(k = 9 \times 10^9\) N m\(^2\) C\(^{-2}\). Positive \(V\) implies positive charge.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Electric potential due to a point charge: \(V = \dfrac{kq}{r}\).
Step 2: Detailed Explanation:
\[ 100 = \frac{9 \times 10^9 \times q}{0.1} \] \[ q = \frac{100 \times 0.1}{9 \times 10^9} = \frac{10}{9 \times 10^9} \approx 1.1 \times 10^{-9} \text{ C} \]
Step 3: Final Answer:
\(q = \mathbf{+1.1 \times 10^{-9}}\) C.