Question

Two positive point charges of 15 x 10-10 C and 13x10-10C are placed
12cm apart. Find the work done in bringing the two charges 4 cm closer.
13
8​

Answer 1

Given:

Two positive point charges of 15 x 10-10 C and 13x10-10C are placed 12cm apart.

To find:

Work done in bringing the two charges 4cm closer ?

Calculation:

The work done for transfer of charges can be calculated from the difference of potential energy at the two distances.

$\sf \therefore \: W = PE2 - PE1$

$\sf \implies\: W = \dfrac{(q1)(q2)}{4\pi \epsilon_{0}(r2)} - \dfrac{(q1)(q2)}{4\pi \epsilon_{0}(r1)}$

$\sf \implies\: W = \dfrac{(q1)(q2)}{4\pi \epsilon_{0}} \bigg \{ \dfrac{1}{r2} - \dfrac{1}{r1} \bigg \}$

Putting the values in SI unit:

$\sf \implies\: W = \dfrac{(15 \times {10}^{ - 10} )(13 \times {10}^{ - 10} )}{4\pi \epsilon_{0}} \bigg \{ \dfrac{1}{ (\frac{8}{100} )} - \dfrac{1}{ (\frac{12}{100} )} \bigg \}$

$\sf \implies\: W = \dfrac{195 \times {10}^{ - 20} }{4\pi \epsilon_{0}} \bigg \{ \dfrac{100}{8} - \dfrac{100}{12} \bigg \}$

$\sf \implies\: W = \dfrac{195 \times {10}^{ - 20} }{4\pi \epsilon_{0}} \bigg \{ \dfrac{300 - 200}{24} \bigg \}$

$\sf \implies\: W = \dfrac{195 \times {10}^{ - 20} }{4\pi \epsilon_{0}} \bigg \{ \dfrac{100}{24} \bigg \}$

$\sf \implies\: W = 195 \times {10}^{ - 20} \times \bigg \{ \dfrac{100}{24} \bigg \} \times 9 \times {10}^{9}$

$\sf \implies\: W = 195 \times {10}^{ - 20} \times \bigg \{ \dfrac{1}{24} \bigg \} \times 9 \times {10}^{11}$

$\sf \implies\: W = 73.12 \times {10}^{ - 9} \: joule$

So, final answer is:

$\boxed{ \bold{\: W = 73.12 \times {10}^{ - 9} \: joule}}$