Question

The work done in bringing a point charge 3 nC from infinity to a point at a distance of 10 cm from a fixed charge 6 nc, is
O 21.8uJ
O 0.2 uJ
O 20 uJ
O 1.62 uJ​

Answer 1

Given:

3 nC charge is brought from infinity to a point at a distance of 10 cm from a fixed charge 6 nC.

To find:

Work done in this process.

Calculation:

Initial Potential Energy of the system:

$\therefore \: PE1 = \dfrac{k(3 \times {10}^{ - 9} \times 6 \times {10}^{ - 9}) }{ \infty}$

$\therefore \: PE1 = 0 \: joule$

Final potential energy of the system:

$\therefore \: PE2 = \dfrac{k(3 \times {10}^{ - 9} \times 6 \times {10}^{ - 9}) }{ 10 \times {10}^{ - 2} }$

$= > \: PE2 = \dfrac{k(18 \times {10}^{ - 18}) }{ {10}^{ - 1} }$

$= > \: PE2 = \dfrac{9 \times {10}^{9} \times (18 \times {10}^{ - 18}) }{ {10}^{ - 1} }$

$= > \: PE2 = \dfrac{162 \times {10}^{ - 9} }{ {10}^{ - 1} }$

$= > \: PE2 = \dfrac{1.62 \times {10}^{ - 7} }{ {10}^{ - 1} }$

$= > \: PE2 = 1.62 \times {10}^{ - 6} \: joule$

So, work done is :

Work = PE2 - PE1

=> Work = 1.6 × 10^(-6) J.

So , final answer is :

$\boxed{ \bold{ \:work = 1.62 \times {10}^{ - 6} \: joule = 1.62 \: \mu J}}$