Question

If two charges q1 and q2 are separated by a distance r, then Potential energy of the system is given by____________ 

1.a

2.b

3.c

4.d

​

Answer 1

Given:

Two charges q1 and q2 are separated by a distance r.

To find:

The value of electrostatic potential energy ?

Diagram:

$\boxed{\setlength{\unitlength}{7mm} \begin{picture}(10,5)\thicklines\put(2,2){\circle*{0.3}} \put(8,2){\circle*{0.3}} \put(2,2){\line(1,0){6}} \put(4,2){\line( - 1,1){0.5}} \put(6,2){\line(1,1){0.5}} \put(4,2){\line( - 1, - 1){0.5}} \put(6,2){\line( 1, -1){0.5}} \put(4.5,0.7){\vector( - 1,0){2.7}} \put(5.4,0.7){\vector(1,0){2.8}}\put(4.75,0.6){ \bf r }\put(2,2.5){q1}\put(8,2.5){q2}\end{picture}}$

Solution:

In this answer, I will try to show you the actual proof with which the expression for potential energy has been derived.

-----------------------------------------------------------

Let consider that the system consists of two charges q1 and q2 ,such that they are separated by distance r.

Now, we can say that ELECTROSTATIC FORCE being a conservative force , so amount of work done to bring a charge from infinity to that separation gets stored in the system as the potential energy.

So, initial potential energy of charge q2 at infinity will be 0 J.

Now, at a seperation of r , potential energy will be equal to the product of charge q2 and the potential provided by q1.

$\therefore \: PE =( V_{q1}) \times q2$

$\implies\: PE = \bigg \{ \dfrac{1}{4\pi \epsilon_{0} } \bigg( \dfrac{q1}{r} \bigg) \bigg \}\times q2$

$\implies\: PE = \dfrac{1}{4\pi \epsilon_{0} } \bigg \{ \dfrac{(q1)(q2)}{r} \bigg \}$

So, the required expression is:

$\boxed{ \bold{\: PE = \dfrac{1}{4\pi \epsilon_{0} } \bigg \{ \dfrac{(q1)(q2)}{r} \bigg \}}}$

Answer 2

Answer:

ahhhhhhh8iggdfhjyrfhitdchhu