Two identical conducting spheres carrying dif | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the initial charges on the two spheres be $Q_1$ and $Q_2$. Since they attract each other,they must have opposite signs,so $Q_1 Q_2 < 0$.The in

Vedclass · Accepted Answer

$ - (3 + \sqrt{8}) $ or $ (-3 + \sqrt{8}) $

Vedclass · Answer

(C) Let the initial charges on the two spheres be $Q_1$ and $Q_2$. Since they attract each other,they must have opposite signs,so $Q_1 Q_2 The initial force is $F = \frac{k |Q_1 Q_2|}{d^2}$. Since they attract,$F = -\frac{k Q_1 Q_2}{d^2}$.When the spheres are brought into contact,the total charge is shared equally because the spheres are identical. The new charge on each sphere is $Q' = \frac{Q_1 + Q_2}{2}$.After returning to their original positions,the new force is $F' = \frac{k (Q')^2}{d^2} = \frac{k (Q_1 + Q_2)^2}{4d^2}$.According to the problem,the magnitude of the new repulsive force is equal to the magnitude of the initial attractive force: $\frac{k (Q_1 + Q_2)^2}{4d^2} = \frac{k |Q_1 Q_2|}{d^2}$.Since $Q_1$ and $Q_2$ have opposite signs,$|Q_1 Q_2| = -Q_1 Q_2$. Thus,$(Q_1 + Q_2)^2 = -4 Q_1 Q_2$.$Q_1^2 + Q_2^2 + 2 Q_1 Q_2 = -4 Q_1 Q_2 \Rightarrow Q_1^2 + Q_2^2 + 6 Q_1 Q_2 = 0$.Dividing by $Q_2^2$,we get $(\frac{Q_1}{Q_2})^2 + 6(\frac{Q_1}{Q_2}) + 1 = 0$.Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,where $x = \frac{Q_1}{Q_2}$:$\frac{Q_1}{Q_2} = \frac{-6 \pm \sqrt{36 - 4}}{2} = \frac{-6 \pm \sqrt{32}}{2} = -3 \pm \sqrt{8}$.

Similar Questions

$A$ charged particle of mass $m_{1}$ and charge $q_{1}$ is revolving in a circle of radius $r$. Another charged particle of charge $q_{2}$ and mass $m_{2}$ is situated at the centre of the circle. If the velocity and time period of the revolving particle are $v$ and $T$ respectively,then:

When two identical point charges are placed at a distance of $5 \, cm$, they experience a repulsive force of $0.144 \, N$. The value of each charge in microcoulombs $(\mu C)$ is:

Three charges are placed as shown in the figure. The magnitude of $q_1$ is $2.00\, \mu C$,but its sign and the value of the charge $q_2$ are not known. Charge $q_3$ is $+4.00\, \mu C$,and the net force on $q_3$ is entirely in the negative $x-$ direction. The magnitude of $q_2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module