JEE MAIN - Physics (2021 - 20th July Morning Shift - No. 8)
A certain charge Q is divided into two parts q and (Q $$-$$ q). How should the charges Q and q be divided so that q and (Q $$-$$ q) placed at a certain distance apart experience maximum electrostatic repulsion?
Q = 2q
Q = 4q
Q = 3q
Q = $${q \over 2}$$
Explanation
Let's say the charge q and (Q $$-$$ q) are at r distance from each other. This can be shown as
_20th_July_Morning_Shift_en_8_1.png)
According to Coulomb's law, force between both the parts can be given as
$$F = {{kq(Q - q)} \over {{r^2}}}$$
$$F = {k \over {{r^2}}}(qQ - {q^2})$$
As we know that $${{dF} \over {dq}} = 0$$, for maximum force.
$$ \Rightarrow {{dF} \over {dq}} = {d \over {dq}}\left[ {{k \over {{r^2}}}(qQ - {q^2})} \right] = 0$$
$$ \Rightarrow {k \over {{r^2}}}(Q - 2q) = 0 \Rightarrow Q = 2q$$
According to Coulomb's law, force between both the parts can be given as
$$F = {{kq(Q - q)} \over {{r^2}}}$$
$$F = {k \over {{r^2}}}(qQ - {q^2})$$
As we know that $${{dF} \over {dq}} = 0$$, for maximum force.
$$ \Rightarrow {{dF} \over {dq}} = {d \over {dq}}\left[ {{k \over {{r^2}}}(qQ - {q^2})} \right] = 0$$
$$ \Rightarrow {k \over {{r^2}}}(Q - 2q) = 0 \Rightarrow Q = 2q$$
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