Question

Two insulated charged copper spheres A and B have their centres separated by a distance of 50cm. The charge on each is \[6.5 \times {10^7}C\]. The radii of A and B are negligible compared to the distance of separation.
Suppose the spheresA and B have identical sizes. A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with the second, finally removed from both. What is the new force of repulsion between A and B?

Answer 1

Hint:Charges flow from higher potential body to lower potential body when they are in contact.Flow of charge will stop when they become equally charged.The electric force (F) acting between two stationary charged particles q1 and q2, separated by distance r, is given by the Coulomb’s law,$F = k\dfrac{{{q_1}{q_2}}}{{{r^2}}}$, where k is coulomb constant.

Complete step by step answer:
Let the charge of the spheres A and B is $Q$.
When the uncharged identical sphere is brought in contact with the\[{1^{st}}\] sphere with charge Q, they will share the charge between them equally and become the same potential.
Now the charge of each sphere\[\left( {{1^{st}}and{\text{ }}the{\text{ }}{3^{rd}}one} \right){\text{ }}is\left( {say} \right)\] .
${q_1} = \dfrac{{Q + 0}}{2} = \dfrac{Q}{2}$
Now the 3rd sphere with charge $\dfrac{Q}{2}$is brought in contact with \[{2^{nd}}\] sphere\[ - B\] with charge $Q$.
They also share the between them.
Let us consider, after sharing the charge between \[{3^{rd}}and{\text{ }}{2^{nd}}\] sphere, each of them will have the same charge ${q_2}$\[\left( {say} \right).\]
${q_2} = \dfrac{{Q + Q/2}}{2} = \dfrac{{3Q}}{4}$
Therefore the sphere B has charge of $\dfrac{{3Q}}{4}$.
Now, the electric force between sphere A and sphere B is given by
$\begin{gathered}
F = k\dfrac{{{q_1}{q_2}}}{{{r^2}}} \\
\Rightarrow F = k\dfrac{{(Q/2) \times (3Q/4)}}{{{r^2}}} = k\dfrac{{3{Q^2}}}{{8{r^2}}} \\
\end{gathered} $
Now in SI unit , $Q$\[ = \] \[6.5 \times {10^7}C\],$k = 9 \times {10^9}(SI)$
$r = 50cm = 0.5m$
\[\begin{gathered}
F =k\dfrac{{3{Q^2}}}{{8{r^2}}} = \dfrac{{9 \times {{10}^9} \times 3 \times {{(6.5 \times {{10}^7})}^2}}}{{8 \times {{0.5}^2}}} = 570.3 \times {10^{23}} \\
\Rightarrow F = 5.7 \times {10^{25}} \\
\end{gathered} \]
The new force of repulsion between A and B \[5.7 \times {10^{25}}\]newton.

Note:Electric force can be attractive or repulsive in nature based on the type of the charge.Same charge repels and opposite charge attracts.