A charge of 4,mu C is to be divided into two | vedclass

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

Question

(B) Let the total charge $Q = 4\,\mu C$ be divided into two parts $q$ and $(Q - q)$.The electrostatic force $F$ between these two charges separated by

Vedclass · Accepted Answer

$2\,\mu C$ and $2\,\mu C$

Vedclass · Answer

(B) Let the total charge $Q = 4\,\mu C$ be divided into two parts $q$ and $(Q - q)$.The electrostatic force $F$ between these two charges separated by a distance $d$ is given by Coulomb's law:$F = \frac{K q (Q - q)}{d^2}$To find the condition for maximum force,we differentiate $F$ with respect to $q$ and set it to zero:$\frac{dF}{dq} = \frac{K}{d^2} \frac{d}{dq} (Qq - q^2) = 0$$\frac{K}{d^2} (Q - 2q) = 0$Since $K$ and $d$ are constants and non-zero,we have:$Q - 2q = 0 \implies q = \frac{Q}{2}$Given $Q = 4\,\mu C$,we get $q = \frac{4\,\mu C}{2} = 2\,\mu C$.Thus,the two charges are $2\,\mu C$ and $2\,\mu C$.

$A$ charge of $4\,\mu C$ is to be divided into two parts. The distance between the two divided charges is constant. What should be the magnitude of the divided charges so that the electrostatic force between them is maximum?

Similar Questions

Two identical spheres are charged with equal charges and the force between them is $F$. If $50\%$ of the charge from one sphere is transferred to the other sphere,then the new force will be ........ $F$.

Two point charges $A$ and $B$,having charges $+Q$ and $-Q$ respectively,are placed at a certain distance apart and the force acting between them is $F$. If $25 \%$ of the charge of $A$ is transferred to $B$,then the force between the charges becomes

Two point charges $q_1$ and $q_2$ are '$l$' distance apart. If one of the charges is doubled and the distance between them is halved, the magnitude of the force becomes $n$ times, where $n$ is:

Two point charges $+Q$ and $+q$ repel each other with a force of $100 \,N$. Keeping the distance between them unchanged, if $Q$ is increased by $10 \%$ and $q$ is decreased by $10 \%$, the force of repulsion between them will:

Vedclass Test Series

Exam Paper Generator

Online Exam Module