Q139 / 180 · KCET 2017 · Physics
Two point charges \( A=+3 \mathrm{nC} \) and \( B=+1 \mathrm{nC} \) are placed \( 5 \mathrm{~cm} \) apart in air. The work done to
move charge \( B \) towards \( A \) by \( 1 \mathrm{~cm} \) is
move charge \( B \) towards \( A \) by \( 1 \mathrm{~cm} \) is
- Correct answerA\( 1.35 \times 10^{-7} \mathrm{~J} \)
- B\( 2.7 \times 10^{-7} \mathrm{~J} \)
- C\( 2.0 \times 10^{-7} \mathrm{~J} \)
- D\( 12.1 \times 10^{-7} \mathrm{~J} \)
Solution
Given
$A=3 n C=3 \times 10^{-9} C ; B=1 n C=1 \times 10^{-9} C$
; distance, $d=5 \mathrm{~cm}=5 \times 10^{-2} \mathrm{~m}$
Therefore, energy of system
$U_{i}=\frac{1}{4 \Pi \varepsilon_{0}} \frac{q_{A} q_{B}}{d}=\left(9 \times 10^{9}\right) \frac{\left(3 \times 10^{-9}\right)\left(1 \times 10^{-9}\right)}{\left(5 \times 10^{-2}\right)}$
To move charge B towards A by $1 \mathrm{~cm}$
$d^{\prime \prime}=4 \mathrm{~cm}=4 \times 10^{-2} \mathrm{~m}$
Therefore, energy of system after moving charge is
$U_{f}=\frac{1}{4 \Pi \varepsilon_{0}} \frac{q_{A} q_{B}}{d}=\frac{\left(9 \times 10^{9}\right)\left(3 \times 10^{-9}\right)\left(1 \times 10^{-9}\right)}{\left(4 \times 10^{-2}\right)}$
Therefore, work done $=U_{f}-U_{i}$
$=\frac{\left(9 \times 10^{9}\right)\left(3 \times 10^{-9}\right)\left(1 \times 10^{-9}\right)}{\left(4 \times 10^{-12}\right)}-\frac{\left(9 \times 10^{9}\right)\left(3 \times 10^{-9}\right)\left(1 \times 10^{-9}\right)}{\left(5 \times 10^{-2}\right)}$
$=\left(9 \times 10^{9}\right)\left(3 \times 10^{-9}\right)\left(1 \times 10^{-9}\right)\left[\frac{1}{4 \times 10^{-2}}-\frac{1}{5 \times 10^{-2}}\right]$
$=\frac{27 \times 10^{-9}}{10^{-2}}\left[\frac{1}{20}\right]=\frac{27}{2} \times 10^{-8}=13.5 \times 10^{-8}$
Thus, work done $=1.35 \times 10^{-7} \mathrm{~J}$
Hence, the correct answer is option: 'A'.