Question

A charge Q is to be divided into two charges Q and Q - Q, such that electrostatic rupulsion between them is maximum at a fixed distance R. The value of Q/Q will​

Answer 1

A charge Q is to be divided into two charges q and Q - q, such that electrostatic repulsion between them is maximum at a fixed distance R. The value of Q/q will​ be:-  

Answer:

• The value of Q/q will be 2

Explanation:

$\rule{300}{1.5}$

There are two charges q and Q - q which are separated by a distance R, Applying coulomb's law of force.

$\longrightarrow\sf F =\dfrac{K\; q (Q-q)}{R^{2}}\\\\\\\\\longrightarrow\sf F =\dfrac{K\; (Qq-q^{2})}{R^{2}}$

As the condition is to be such that electronic repulsion b/w them should be maximum, For this condition dF/dq should be 0.

$\longrightarrow\sf \dfrac{dF}{dq}=0\\\\\\\\\longrightarrow\sf \dfrac{d}{dq}\Bigg(\dfrac{K\;(Qq-q^{2})}{R^{2}}\Bigg)=0\\\\\\\\\longrightarrow\sf \dfrac{K\;(Q-2q)}{R^{2}}=0\\\\\\\\\longrightarrow\sf Q-2q=0\\\\\\\\\longrightarrow\sf Q=2q\\\\\\\\\longrightarrow\large{\underline{\boxed{\sf \dfrac{Q}{q}=2}}}$

∴ The value of Q/q will be 2.

$\rule{300}{1.5}$

Answer 2

Answer:

The value of Q/q is 2

Explanation:

Given: Magnitude of the two charges $q$ and $Q-q$ respectively.

Distance between them = $R$

Now, Coulombic repulsion between charges is given by,

$F=\frac{Kq(Q-q)}{R^2}$

$F=\frac{KqQ}{R^2}-\frac{Kq^2}{R^2}$

For maximum force, $\frac{dF}{dq}$ must be eqal to zero.

Now, $\frac{dF}{dq} = \frac{Kq}{R^2}-\frac{2Kq}{R^2}$

equating it to zero,

$\implies \frac{dF}{dq} =0= \frac{K(Q-2q)}{R^2}$

$\implies 0=(Q-2q)$

$\implies Q=2q$

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

The value of Q/q is 2.

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