Answer:
Explanation:
Given
Radius of Conducting Sphere is R with Positive charge q
initially small sphere is at a distance of r=5 R and slowly move to r=3 R
Electric Potential Energy Between two Charged Particle
[tex]U=k\cdot \frac{q_1q_2}{r^2}[/tex]
Initial Potential Energy
[tex]U_1=k\cdot \frac{qq_0}{(5R)^2}[/tex]
[tex]U_1=k\cdot \frac{qq_0}{25R^2}[/tex]
at r=3R
[tex]U_2=k\cdot \frac{qq_0}{(3R)^2}[/tex]
[tex]U_2=k\cdot \frac{qq_0}{9R^2}[/tex]
Work Done [tex]=U_2-U_1[/tex]
[tex]W=k\cdot \frac{qq_0}{9R^2}-k\cdot \frac{qq_0}{25R^2}[/tex]
[tex]W=k\cdot \frac{qq_0}{R^2}\left [ \frac{1}{9}-\frac{1}{25}\right ][/tex]
[tex]W=k\cdot \frac{qq_0}{R^2}\cdot \left [ \frac{16}{225}\right ][/tex]
[tex]W=\frac{1}{4\pi \epsilon _0}\cdot \frac{qq_0}{R^2}\cdot \left [ \frac{16}{225}\right ][/tex]
[tex]W=\frac{4qq_0}{225\pi \epsilon _0R^2}[/tex]