A charge q is placed at the centre of the lin | vedclass

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

Question

(B) Let the two charges $Q$ be placed at points $A$ and $B$ separated by a distance $2r$. The charge $q$ is placed at the midpoint $O$.For the system

Vedclass · Accepted Answer

$-\frac{Q}{4}$

Vedclass · Answer

(B) Let the two charges $Q$ be placed at points $A$ and $B$ separated by a distance $2r$. The charge $q$ is placed at the midpoint $O$.For the system to be in equilibrium,the net force on each charge must be zero.Consider the force on charge $Q$ at point $A$:The force due to charge $q$ at $O$ is $F_1 = \frac{1}{4 \pi \varepsilon_0} \frac{Qq}{r^2}$ (attractive,towards $O$).The force due to charge $Q$ at $B$ is $F_2 = \frac{1}{4 \pi \varepsilon_0} \frac{Q^2}{(2r)^2} = \frac{1}{4 \pi \varepsilon_0} \frac{Q^2}{4r^2}$ (repulsive,away from $B$).For equilibrium,$F_1 + F_2 = 0$,so $\frac{1}{4 \pi \varepsilon_0} \frac{Qq}{r^2} + \frac{1}{4 \pi \varepsilon_0} \frac{Q^2}{4r^2} = 0$.Solving for $q$: $\frac{Qq}{r^2} = -\frac{Q^2}{4r^2} \Rightarrow q = -\frac{Q}{4}$.

$A$ charge $q$ is placed at the centre of the line joining two equal charges $Q$. The system of the three charges will be in equilibrium if $q$ is equal to:

Similar Questions

In almost all Indian metropolitan cities like Delhi,the major atmospheric pollutant$(s)$ is/are:

Which type of chromosome is shown in the figure below?

If $p$-th,$q$-th,and $r$-th terms of a geometric progression are the positive numbers $a, b,$ and $c$ respectively,then the angle between the vectors $(\log a^2) i + (\log b^2) j + (\log c^2) k$ and $(q-r) i + (r-p) j + (p-q) k$ is

If the $x$-coordinate of a point $P$ on the line joining the points $Q(2, 2, 1)$ and $R(5, 2, -2)$ is $4$,then the $z$-coordinate of $P$ is:

If $x$ is real,then the maximum and minimum values of the expression $\frac{x^2-3x+4}{x^2+3x+4}$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module