Two identical charged conducting spheres A an | vedclass

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

Question

(D) Initially,the force between spheres $A$ and $B$ is given by Coulomb's law: $F = \frac{k q^2}{r^2}$.
When the third identical uncharged sphere

Vedclass · Accepted Answer

$\frac{3 F}{8}$

Vedclass · Answer

(D) Initially,the force between spheres $A$ and $B$ is given by Coulomb's law: $F = \frac{k q^2}{r^2}$.
When the third identical uncharged sphere $C$ is brought in contact with sphere $A$,the total charge $q + 0 = q$ is shared equally between them. Thus,the new charge on $A$ is $q_A' = \frac{q}{2}$ and on $C$ is $q_C' = \frac{q}{2}$.
Next,sphere $C$ (now with charge $\frac{q}{2}$) is brought in contact with sphere $B$ (which has charge $q$). The total charge is $q + \frac{q}{2} = \frac{3q}{2}$. This charge is shared equally between $B$ and $C$. Thus,the new charge on $B$ is $q_B' = \frac{3q}{4}$ and on $C$ is $q_C'' = \frac{3q}{4}$.
The new force of repulsion between $A$ and $B$ is $F' = \frac{k q_A' q_B'}{r^2} = \frac{k (q/2) (3q/4)}{r^2} = \frac{3}{8} \frac{k q^2}{r^2}$.
Since $F = \frac{k q^2}{r^2}$,we get $F' = \frac{3 F}{8}$.

Similar Questions

Three equal charges are placed on the three corners of a square. If the force between $q_1$ and $q_2$ is $F_{12}$ and that between $q_1$ and $q_3$ is $F_{13}$,the ratio of magnitudes $\frac{F_{12}}{F_{13}}$ is

What should be the equal charge on helium-filled balloons as shown in the figure?

Two unit negative charges are placed on a straight line. $A$ positive charge $q$ is placed exactly at the midpoint between these unit charges. If the system of these three charges is in equilibrium,the value of $q$ (in $C$) is

Two point charges $+q_{1}$ and $+q_{2}$ are placed a finite distance '$d$' apart. It is desired to put a third charge $q_{3}$ in between these two charges so that $q_{3}$ is in equilibrium. This is

Vedclass Test Series

Exam Paper Generator

Online Exam Module