Two identical tennis balls each having mass m | vedclass

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

Question

(C) Let $T$ be the tension in the thread and $x$ be the separation between the balls.At equilibrium,the forces acting on each ball are: tension $T$,we

Vedclass · Accepted Answer

$x=\left(\frac{q^{2} l}{2 \pi \varepsilon_{0} mg}\right)^{1 / 3}$

Vedclass · Answer

(C) Let $T$ be the tension in the thread and $x$ be the separation between the balls.At equilibrium,the forces acting on each ball are: tension $T$,weight $mg$,and electrostatic force $F_e = \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{x^2}$.Resolving the forces,we have:$T \cos 	heta = mg$ (vertical component)$T \sin 	heta = \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{x^2}$ (horizontal component)Dividing the two equations: $	an 	heta = \frac{q^2}{4 \pi \varepsilon_0 x^2 mg}$.For small angles,$	an 	heta \approx \sin 	heta = \frac{x/2}{l} = \frac{x}{2l}$.Substituting this into the equation: $\frac{x}{2l} = \frac{q^2}{4 \pi \varepsilon_0 x^2 mg}$.Rearranging for $x^3$: $x^3 = \frac{q^2 l}{2 \pi \varepsilon_0 mg}$.Thus,$x = \left(\frac{q^2 l}{2 \pi \varepsilon_0 mg}ight)^{1/3}$.

Two identical tennis balls each having mass $m$ and charge $q$ are suspended from a fixed point by threads of length $l$. What is the equilibrium separation when each thread makes a small angle $\theta$ with the vertical?

Similar Questions

$A$ charge of $1 \mu C$ is divided into two parts such that their charges are in the ratio of $2: 3$. These two charges are kept at a distance of $1 \ m$ apart in vacuum. Then,the electric force between them (in $N$) is

Point charges $+4q, -q$ and $+4q$ are kept on the $x$-axis at points $x = 0, x = a$ and $x = 2a$ respectively. Then:

$A$ 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:

Three charges $+3q$,$Q$,and $+q$ are placed in a straight line of length $\ell$ at points at distances $0$,$\frac{\ell}{2}$,and $\ell$ respectively. The value of $Q$ in order to have the net force on $+q$ to be zero is $Q = xq$. The value of $x$ is:

Two positive point charges are separated by a distance of $4 \ m$ in air. If the sum of the two charges is $36 \mu C$ and the electrostatic force between them is $0.18 \ N$,then the bigger charge is (in $\mu C$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module