Two charges of equal magnitudes and at a dist | vedclass

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

Question

(D) According to Coulomb's Law,the force between two charges $Q_1$ and $Q_2$ separated by a distance $r$ is given by $F = k \frac{Q_1 Q_2}{r^2}$.Given

Vedclass · Accepted Answer

$F / 16$

Vedclass · Answer

(D) According to Coulomb's Law,the force between two charges $Q_1$ and $Q_2$ separated by a distance $r$ is given by $F = k \frac{Q_1 Q_2}{r^2}$.Given that the charges are equal in magnitude,let $Q_1 = Q_2 = Q$. Thus,the initial force is $F = k \frac{Q^2}{r^2}$.When the charges are halved,the new charges become $Q' = Q/2$. When the distance is doubled,the new distance becomes $r' = 2r$.The new force $F'$ is given by $F' = k \frac{(Q/2)(Q/2)}{(2r)^2}$.$F' = k \frac{Q^2 / 4}{4r^2} = \frac{1}{16} \left( k \frac{Q^2}{r^2} \right)$.Since $F = k \frac{Q^2}{r^2}$,we have $F' = F / 16$.

Two charges of equal magnitudes and at a distance $r$ exert a force $F$ on each other. If the charges are halved and the distance between them is doubled,then the new force acting on each charge is

Similar Questions

Three charges are placed at the vertices of an equilateral triangle of side '$a$' as shown in the figure. The force experienced by the charge placed at the vertex $A$ in a direction normal to $BC$ is

In the given diagram,find the distance of the neutral point from the particle of charge $e$ in $cm$.

By using Coulomb's law,define unit charge.

$F$ is the force between two identical charged particles placed at a distance $Y$ from each other. If the distance between the charges is reduced to half the previous distance,then the force between them becomes:

Why is the Coulombian force called a two-body force?

Vedclass Test Series

Exam Paper Generator

Online Exam Module