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GUJCET · Physics · Electrostatic Potential and Capacitance
Charges \(5 \mu C\) and \(10 \mu C\) are placed 1 m apart. Work done to bring these charges at a distance 0.5 m from each other is __________ .
\(\left( k =9 \times 10^9 SI \right)\)
- A \(9 \times 10^4 J\)
- B \(18 \times 10^4 J\)
- C \(45 \times 10^{-2} J\)
- D \(9 \times 10^{-1} J\)
Answer & Solution
Correct Answer
(C) \(45 \times 10^{-2} J\)
Step-by-step Solution
Detailed explanation
(C)
\(\begin{array}{l} W =\Delta U = U _2- U _1 \\ W=\frac{k q_1 q_2}{r_2}-\frac{k q_1 q_2}{r_1} \\ W=k q_1 q_2\left(\frac{1}{r_2}-\frac{1}{r_1}\right)\end{array}\)
\(\begin{array}{r}\therefore \quad W=9 \times 10^9 \times 5 \times 10^{-6} \times 10 \\ \times 10^{-6}\left[\frac{1}{0.5}-\frac{1}{1}\right]\end{array}\)
\(\therefore \quad W =45 \times 10^{-2} J\)
\(\begin{array}{l} W =\Delta U = U _2- U _1 \\ W=\frac{k q_1 q_2}{r_2}-\frac{k q_1 q_2}{r_1} \\ W=k q_1 q_2\left(\frac{1}{r_2}-\frac{1}{r_1}\right)\end{array}\)
\(\begin{array}{r}\therefore \quad W=9 \times 10^9 \times 5 \times 10^{-6} \times 10 \\ \times 10^{-6}\left[\frac{1}{0.5}-\frac{1}{1}\right]\end{array}\)
\(\therefore \quad W =45 \times 10^{-2} J\)
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