Three point charges +q,+2q,and +4q are placed | vedclass

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

Question

(D) Let the distance between $+q$ and $+2q$ be $r$,and the distance between $+2q$ and $+4q$ be $r$. Thus,the distance between $+q$ and $+4q$ is $2r$.T

Vedclass · Accepted Answer

$1: 3$

Vedclass · Answer

(D) Let the distance between $+q$ and $+2q$ be $r$,and the distance between $+2q$ and $+4q$ be $r$. Thus,the distance between $+q$ and $+4q$ is $2r$.The net electrostatic force $F_1$ on charge $+q$ is the sum of forces due to $+2q$ and $+4q$:$F_1 = \frac{k(q)(2q)}{r^2} + \frac{k(q)(4q)}{(2r)^2} = \frac{2kq^2}{r^2} + \frac{4kq^2}{4r^2} = \frac{2kq^2}{r^2} + \frac{kq^2}{r^2} = \frac{3kq^2}{r^2}$.The net electrostatic force $F_2$ on charge $+4q$ is the sum of forces due to $+2q$ and $+q$:$F_2 = \frac{k(4q)(2q)}{r^2} + \frac{k(4q)(q)}{(2r)^2} = \frac{8kq^2}{r^2} + \frac{4kq^2}{4r^2} = \frac{8kq^2}{r^2} + \frac{kq^2}{r^2} = \frac{9kq^2}{r^2}$.The ratio of the net electrostatic force on charges $+q$ and $+4q$ is:$\frac{F_1}{F_2} = \frac{3kq^2/r^2}{9kq^2/r^2} = \frac{3}{9} = 1:3$.

Three point charges $+q$,$+2q$,and $+4q$ are placed along a straight line such that the charge $+2q$ is equidistant from the other two charges. The ratio of the net electrostatic force on charges $+q$ and $+4q$ is

Similar Questions

Two charges each equal to $2\,\mu C$ are $0.5\,m$ apart. If both of them exist inside vacuum,then the force between them is.......$N$

$A$ charge $Q$ is divided into two parts $Q_1$ and $Q_2$. These charges are placed at a distance $R$. What will be the values of $Q_1$ and $Q_2$ for the maximum repulsive force between them?

$A$ 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:

The electrical force between two point charges is $200 \ N$. If we increase $10 \ \%$ charge on one of the charges and decrease $10 \ \%$ charge on the other,then the electrical force between them for the same distance becomes . . . . . . . . . . (in $N$)

Two equal charges are separated by a distance $d$. $A$ third charge placed on a perpendicular bisector at $x$ distance will experience maximum Coulomb force when

Vedclass Test Series

Exam Paper Generator

Online Exam Module