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?

Question

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? (A) Q=(q/2) (B) Q=2 q (C) Q=4 q (D) Q=3 q

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/2c078924da309bb35e9524e2fe5d9c66-.png" /> Fq=(k q(Q-q)/L2)=(k/L2)(q Q-q2) (d F/d q)=0 when force is maximum (d F/d q)=(k/L2)[Q-2 q]=0 ⇒ Q-2 q=0 ⇒ Q=2 q

Q. 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?

$Q = \frac{q}{2}$

$Q = 2 q$

$Q = 4 q$

$Q = 3 q$

$F_{q} = \frac{k q ( Q - q )}{L ^{2}} = \frac{k}{L ^{2}} (qQ - q^{2})$
$\frac{d F}{d q} = 0$ when force is maximum
$\frac{d F}{d q} = \frac{k}{L ^{2}} [Q - 2 q] = 0$
$\Rightarrow Q - 2 q = 0$
$\Rightarrow Q = 2 q$

Q. 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=2q

Q=4q

Q=3q