Solveeit Logo
Login

Question

Physics Question on Electrostatic potential

Two point charges A=+3 nCA = +3\, nC and B=+1 nCB = +1\, nC are placed 5 cm5\,cm apart in air. The work done to move charge BB towards AA by 1 cm1\,cm is

A

1.35×10−7J1.35 \times 10^{-7} J

B

2.7×10−7J2.7 \times 10^{-7} J

C

2.0×10−7J2.0 \times 10^{-7} J

D

12.1×10−7J12.1 \times 10^{-7} J

Answer

1.35×10−7J1.35 \times 10^{-7} J

Explanation

Solution

Given,
A=+3nC=3×10−9 CA=+3 nC =3 \times 10^{-9} \,C
B=+1nC=1×10−9CB=+1 nC =1 \times 10^{-9} C
Distance, I1=5 cm=0.05 m=5×10−2 m I_{1}=5 \,cm =0.05 \,m =5 \times 10^{-2} \,m
Work done (W)=UB−UA(W)=U_{B}-U_{A}
=kq1q2I2−kq1q2I1=\frac{k q_{1} q_{2}}{I_{2}}-\frac{k q_{1} q_{2}}{I_{1}}
Here, I2=I1−1I_{2}=I_{1}-1
I2=5−1=4 cm=0.04 m=4×10−2 mI_{2}=5-1=4\, cm =0.04 \,m =4 \times 10^{-2} \,m
=kq1q2[1I2−1r1]= k q_{1} q_{2}\left[\frac{1}{I_{2}}-\frac{1}{r_{1}}\right]
=9×109×3×10−9×1×10−9=9 \times 10^{9} \times 3 \times 10^{-9} \times 1 \times 10^{-9}
[14×10−2−15×10−2]\left[\frac{1}{4 \times 10^{-2}}-\frac{1}{5 \times 10^{-2}}\right]
=27×10−9×15×4×10−2= \frac{27 \times 10^{-9} \times 1}{5 \times 4 \times 10^{-2}}
=2720×10−7=\frac{27}{20} \times 10^{-7}
=135×10−7 J=135 \times 10^{-7} \,J