A charge having magnitude Q is divided into two parts q and (Q-q) which are held a certian distance r apart. The force of repulsion between the two parts will be maximum if the ratio $\frac{q}{Q}$ is

Q. A certain charge

Q

is to be divided into two parts

q

and

Q - q

. The relationship between

Q

and

q

placed at a certain distance apart to have the maximum Coulomb repulsion is

Q. A certain charge

Q

is divided at first into two parts,

(q)

and

(Q - q)

. Later on the chrages are placed at a certain distance. If the force of interaction between the two charges is maximum, then

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A charge Q is divided into two parts. The two charges kept at a distance apart have a maximum columbian repulsion. Then the ratio of Q and one of the parts is given by :

The correct option is B 2:1Total charge is QNow let it is divided into q and (Q-q)Force of repulsion is F=q(Q−q)r2For F to be maximum dFdq=0⇒Q−qr2−qr2=0Qq=2

The correct option is B $2 : 1$
Total charge is Q
Now let it is divided into q and (Q-q)
Force of repulsion is $F = \frac{q (Q - q)}{r^{2}}$
For F to be maximum $\frac{d F}{d q} = 0$
$\Rightarrow \frac{Q - q}{r^{2}} - \frac{q}{r^{2}} = 0$
$\frac{Q}{q} = 2$