Two positive point charges of 12 and 5 microcoulombs, are placed 10 cm apart in air. The work needed to bring them 4 cm closer is

Options

A. 2.4 J
B. 3.6 J
C. 1.6 J
D. 6.0 J

Answer

B

Explanation

Potential energy of charges `q_(1) and q_(2)`, r distance aprat
`U=(1)/(4pi epsilon_(0))(q_(1)q_(2))/(r)`
For r = 0.1 m
`U_(1)=(1)/(4pi epsilon_(0))(12xx10^(-6)xx5xx10^(-6))/(0.1)`
`=(9xx10^(9)xx60xx10^(-12))/(0.1)=5.4J`
For r = 0.06 m,
`U_(2)=(9xx10^(9)xx60xx10^(-12))/(0.06)=9J`
`therefore" Work done"=(9-5.4)J=3.6J`