Two identical conducting spheres P and S with | vedclass

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

Question

(B) Initially, the force between spheres $P$ and $S$ is given by Coulomb's law: $F = k \frac{Q^2}{r^2} = 16 \,N$.When sphere $R$ (uncharged) is brough

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(B) Initially, the force between spheres $P$ and $S$ is given by Coulomb's law: $F = k \frac{Q^2}{r^2} = 16 \,N$.When sphere $R$ (uncharged) is brought in contact with sphere $P$ (charge $Q$), the charge is shared equally between them because they are identical. Thus, the new charge on $P$ is $Q_P' = \frac{Q+0}{2} = \frac{Q}{2}$. The charge on $R$ becomes $\frac{Q}{2}$.Next, sphere $R$ (now with charge $\frac{Q}{2}$) is brought in contact with sphere $S$ (charge $Q$). The total charge is shared equally between them. Thus, the new charge on $S$ is $Q_S' = \frac{Q + Q/2}{2} = \frac{3Q/2}{2} = \frac{3Q}{4}$.The new force of repulsion between $P$ and $S$ is $F' = k \frac{Q_P' Q_S'}{r^2} = k \frac{(Q/2)(3Q/4)}{r^2} = \frac{3}{8} \left( k \frac{Q^2}{r^2} ight)$.Substituting the initial force $F = 16 \,N$, we get $F' = \frac{3}{8} 	imes 16 \,N = 6 \,N$.

Two identical conducting spheres $P$ and $S$ with charge $Q$ on each, repel each other with a force $16 \,N$. $A$ third identical uncharged conducting sphere $R$ is successively brought in contact with the two spheres. The new force of repulsion between $P$ and $S$ is: (in $\,N$)

Similar Questions

Two point charges $+9e$ and $+e$ are at $16\, cm$ away from each other. Where should another charge $q$ be placed between them so that the system remains in equilibrium?

Equal charges $q$ are placed at the four corners $A, B, C, D$ of a square of side length $a$. The magnitude of the net electrostatic force on the charge at $B$ is:

Two particles of equal mass $m$ and charge $q$ are placed at a distance of $16 \, cm$. They do not experience any net force. The value of $\frac{q}{m}$ is

$A$ proton of mass $m$ and charge $e$ is projected from a very large distance towards an $\alpha$-particle with velocity $v$. Initially,the $\alpha$-particle is at rest,but it is free to move. If gravity is neglected,then the minimum separation along the straight line of their motion will be:

There are two charges $+1 \ \mu C$ and $+5 \ \mu C$. The ratio of the forces acting on them will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module