A charge Q is divided into two parts of q and Q-q. If the coulomb repulsion between them when they are separated is to be maximum, the ratio of (Q/q) should be

Question

A charge Q is divided into two parts of q and Q-q. If the coulomb repulsion between them when they are separated is to be maximum, the ratio of (Q/q) should be (A) 2 (B) 1/2 (C) 4 (D) 1/4

Tardigrade Admin · Accepted Answer

Let distance between two charges is x The force between them. F=(1/4 π ε0) (q(Q-q)/x2) x is constant, so for maximum force q(Q-q) should be maximum. ( d / L q) q(Q-q) =0 Q-2 q=0 ⇒ (Q/q) =2

Q. A charge $Q$ is divided into two parts of $q$ and $Q - q$ . If the coulomb repulsion between them when they are separated is to be maximum, the ratio of $\frac{Q}{q}$ should be

2

1/2

4

1/4

Let distance between two charges is $x$
The force between them.
$F = \frac{1}{4 π ε _{0}} \frac{q ( Q - q )}{x ^{2}}$
$x$ is constant, so for maximum force $q (Q - q)$ should be maximum.
$\frac{d}{L q} {q (Q - q)} = 0$
$Q - 2 q = 0$
$\Rightarrow \frac{Q}{q} = 2$

Q. A charge Q is divided into two parts of q and Q−q. If the coulomb repulsion between them when they are separated is to be maximum, the ratio of qQ​ should be

2

1/2

4

1/4

Let distance between two charges is x The force between them. F=4πε0​1​x2q(Q−q)​ x is constant, so for maximum force q(Q−q) should be maximum. Lqd​{q(Q−q)}=0 Q−2q=0 ⇒qQ​=2

Q. A charge $Q$ is divided into two parts of $q$ and $Q - q$ . If the coulomb repulsion between them when they are separated is to be maximum, the ratio of $\frac{Q}{q}$ should be

Let distance between two charges is $x$
The force between them.
$F = \frac{1}{4 π ε _{0}} \frac{q ( Q - q )}{x ^{2}}$
$x$ is constant, so for maximum force $q (Q - q)$ should be maximum.
$\frac{d}{L q} {q (Q - q)} = 0$
$Q - 2 q = 0$
$\Rightarrow \frac{Q}{q} = 2$