Question

Two point charges $A=+3$ nC and $B=+1$ nC are placed $5cm$ apart in air. The work done to move charge $B$ towards $A$ by $1cm$ is :

• A. $2.0\times {10}^{-7}J$
• B. $1.35\times {10}^{-7}J$
• C. $2.7\times {10}^{-7}J$
• D. $12.1\times {10}^{-7}J$

Answer 1

Answer: (B)

$1.35\times {10}^{-7}J$

Initial distance between the charges $Q_A$ and $Q_B$, $d=5$ cm $=0.05$ m

Potential energy of the system initially, $U_i=\frac{k{Q}_A{Q}_B}{d}$ where $k=9\times {10}^9$

$\therefore$ $U_i=\frac{9\times {10}^9\times 3\times {10}^{-9}\times 1\times {10}^{-9}}{0.05}=5.4\times {10}^{-7}J$

Now the charge $B$ is moved by $1cm$ towards $A$.

Thus new distance between the charges $d^{\prime }=5-1=4$ cm $=0.04$ m

Potential energy of the system initially, $U_f=\frac{k{Q}_A{Q}_B}{d^{\prime }}$

$\therefore$ $U_f=\frac{9\times {10}^9\times 3\times {10}^{-9}\times 1\times {10}^{-9}}{0.04}=6.75\times {10}^{-7}J$

Thus work done $W=U_f-U_i$

$\therefore$ $W=(6.75-5.4)\times {10}^{-7}=1.35\times {10}^{-7}J$