Charge Q is divided into two parts which are then kept some distance apart. The force between them will be maximum if the two parts are

Question

Charge Q is divided into two parts which are then kept some distance apart. The force between them will be maximum if the two parts are (A) Q / 2 each (B) Q / 4 and 3 Q / 4 (C) Q / 3 and 2 Q / 3 (D) e and (Q-e), where e= electronic charge.

Tardigrade Admin · Accepted Answer

F=((1/4 π ε0)) ⋅((q( Q -q)/r2)) F is maximum, when q(Q-q) is maximum. F prime(q)=Q-2 q For maxima, F prime(q)=Q-2 q=0 or q=(Q/2) F prime prime(q)=-2<0 (Maxima).

Q. Charge $Q$ is divided into two parts which are then kept some distance apart. The force between them will be maximum if the two parts are

$Q /2$ each

$Q /4$ and $3 Q /4$

$Q /3$ and $2 Q /3$

$e$ and $(Q - e),$ where $e =$ electronic charge.

$F = (\frac{1}{4 π ε _{0}}) \cdot (\frac{q ( Q - q )}{r ^{2}})$
$F$ is maximum, when $q (Q - q)$ is maximum.
$F^{'} (q) = Q - 2 q$
For maxima, $F^{'} (q) = Q - 2 q = 0$
or $q = \frac{Q}{2}$
$F^{''} (q) = - 2 < 0$ (Maxima).

Q. Charge Q is divided into two parts which are then kept some distance apart. The force between them will be maximum if the two parts are

Q/2 each

Q/4 and 3Q/4

Q/3 and 2Q/3

e and (Q−e), where e= electronic charge.