Two charges q1 and q2 are separated bya distance d. calculate the work

1.	Two charges q1 and q2 are separated bya distance d. calculate the work done inincreasing the distance to 2d
Answer» Given: Two charges q1 and Q2 are separated by a distance d. To find: Work done in increasing the distance upto 2d. Calculation: Initial POTENTIAL ENERGY between the two charges $U1 = \dfrac{1}{4\pi \epsilon_{0}} \bigg \{ \dfrac{(q1)(q2)}{d} \bigg \}$ After changing the separation distance to 2d , new Potential Energy: $U2 = \dfrac{1}{4\pi \epsilon_{0}} \bigg \{ \dfrac{(q1)(q2)}{2d} \bigg \}$ So, work done will be equal to change in Potential Energy of the system: $\therefore \: work = U2 - U1$ $= > work = \dfrac{1}{4\pi \epsilon_{0}} \bigg \{ \dfrac{(q1)(q2)}{2d} \bigg \} - \dfrac{1}{4\pi \epsilon_{0}} \bigg \{ \dfrac{(q1)(q2)}{d} \bigg \}$ $= > work = - \dfrac{1}{4\pi \epsilon_{0}} \bigg \{ \dfrac{(q1)(q2)}{2d} \bigg \}$ So, final answer is: $\boxed{ \sf{work = - \dfrac{1}{4\pi \epsilon_{0}} \bigg \{ \dfrac{(q1)(q2)}{2d} \bigg \} }}$ [SINCE work done is negative, it signifies the fact that it the change in distance between the charged system will be spontaneous.]

Two charges q1 and q2 are separated bya distance d. calculate the work done inincreasing the distance to 2d

Answer»

Two charges q1 and Q2 are separated by

a distance d.

Work done in increasing the distance upto 2d.

Initial POTENTIAL ENERGY between the two charges

$U1 = \dfrac{1}{4\pi \epsilon_{0}} \bigg \{ \dfrac{(q1)(q2)}{d} \bigg \}$

After changing the separation distance to 2d , new Potential Energy:

$U2 = \dfrac{1}{4\pi \epsilon_{0}} \bigg \{ \dfrac{(q1)(q2)}{2d} \bigg \}$

So, work done will be equal to change in Potential Energy of the system:

$\therefore \: work = U2 - U1$

$= > work = \dfrac{1}{4\pi \epsilon_{0}} \bigg \{ \dfrac{(q1)(q2)}{2d} \bigg \} - \dfrac{1}{4\pi \epsilon_{0}} \bigg \{ \dfrac{(q1)(q2)}{d} \bigg \}$

$= > work = - \dfrac{1}{4\pi \epsilon_{0}} \bigg \{ \dfrac{(q1)(q2)}{2d} \bigg \}$

So, final answer is:

$\boxed{ \sf{work = - \dfrac{1}{4\pi \epsilon_{0}} \bigg \{ \dfrac{(q1)(q2)}{2d} \bigg \} }}$

[SINCE work done is negative, it signifies the fact that it the change in distance between the charged system will be spontaneous.]