Two positive point charges of 12 and 8 microcoulombs respectively are placed 10 cm apart in air. The work done to bring them 4 cm closer is

A
zero
B
3.8 J
C
4.8 J
D
5.8 J

Solution:

Electrostatic potential energy when the charges are 10 cm = 0.1 m apart is

$U_{1} = \frac{q_{1} q_{2}}{4 {πε}_{0} r} = \frac{12 \times 10^{- 6} \times 8 \times 10^{- 6}}{4 {πε}_{0} \times 0.1} = \frac{9.6 \times 10^{- 10}}{4 {πε}_{0}}$

Potential energy when the charges are brought 4 cm closer, i.e., when they are 6 cm = 0.06 m apart is

$U_{2} = \frac{12 \times 10^{- 6} \times 8 \times 10^{- 6}}{4 {πε}_{0} \times 0.06} = \frac{16 \times 10^{- 10}}{4 {πε}_{0}}$

$∴ W o r k d o n e = U_{2} - U_{1} = \frac{10^{- 10}}{4 {πε}_{0}} (16 - 9.6) = 9 \times 10^{9} \times 10^{- 10} \times 6.4 = 5.76 J = 5.8 J$

Also Check:

Two positive point charges of 12 and 8 microcoulombs respectively are placed 10 cm apart in air. The work done to bring them 4 cm closer is

Solution:

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