Electrostatics> Two identical spheres A and B are equally...

Two identical spheres A and B are equally charged and seperated by a certain distance. Another identical uncharged sphere C is brought in contact with A and then placed mid way between A and B,. Calculate the resultant force on C if initial force between A and B is 2*10^-5 N.

Sanjay , 7 Years ago

Grade 12

2 Answers

Eshan

Last Activity: 7 Years ago

Let the distance between the charges be $r$ and their charges be $q$ each.

Hence the force between them is $\dfrac{kq^2}{r^2}=2\times 10^{-5}N$

When identical uncharged sphere is brought in contact, the charge on the charged sphere gets divided equally. Hence charge on both A and C becomes $\dfrac{q}{2}$ .
Hence force on C due to A after it is moved to the mid point= $\dfrac{k(\dfrac{q}{2})(\dfrac{q}{2})}{(\dfrac{r}{2})^2}=\dfrac{kq^2}{r^2}=F$

Force on C due to B= $\dfrac{kq(\dfrac{q}{2})}{(\dfrac{r}{2})^2}=\dfrac{2kq^2}{r^2}=2F$
Hence net force on C=2F-F=F

Gitanjali Rout

Last Activity: 7 Years ago

Hence net force on C=2F-F=F $\dfrac{kq(\dfrac{q}{2})}{(\dfrac{r}{2})^2}=\dfrac{2kq^2}{r^2}=2F$ Force on C due to B= $\dfrac{k(\dfrac{q}{2})(\dfrac{q}{2})}{(\dfrac{r}{2})^2}=\dfrac{kq^2}{r^2}=F$ .Hence force on C due to A after it is moved to the mid point= $\dfrac{q}{2}$ When identical uncharged sphere is brought in contact, the charge on the charged sphere gets divided equally. Hence charge on both A and C becomes $\dfrac{kq^2}{r^2}=2\times 10^{-5}N$ each.Hence the force between them is $q$ and their charges be $r$ Let the distance between the charges be

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