Question

EXAMPLE 10. Two balls each of mass 0.1 mg carry
identical charges and suspended by two non-conducting
threads of same length (0.4 m). In equilibrium, each thread
makes an angle of 30° with the vertical. Find charge on each
ball.​

Answer 1

Given :

• Mass of balls,$\sf\:m=0.1mg=0.1\times10^{-6}$
• Length of thread , l = 0.4 m
• Angle made by threads with vertical at equilibrium,θ = 30°

To Find :

Charge on each ball

Solution :

Let the charge on each ball be " q " .

Let's Understand the problem , Draw free body diagram of given condition .

From the " Free body Diagram "

At equilibrium condition :

$\sf\:T\:\cos\theta=mg...(1)$

and $\sf\:T\:\sin\theta=F_{q}.....(2)$

Distance , Between balls , r = 2l sinθ

By columb's Law

$\sf\:F_q=\dfrac{K\:q^2}{r^2}$

Put the given values :

$\sf\implies\:F_q=\dfrac{9\times10^9\times\:q^2}{(2l\sin\:30\degree)^2}$

$\sf\implies\:F_q=\dfrac{9\times10^9\times\:q^2}{l^2}$

$\sf\implies\:F_q=\dfrac{9\times10^9\times\:q^2}{(0.4)^2}$

$\sf\implies\:F_q=\dfrac{9\times10^9\times\:q^2}{0.16}$

$\sf\implies\:F_q=\dfrac{9\times10^{11}\times\:q^2}{16}$

Now , Divide Equation (2) and (1) , Then

$\sf\dfrac{T\:\sin\theta}{T\:\cos\theta}=\dfrac{F_q}{mg}$

$\sf\implies\tan\:30\degree=\dfrac{\frac{9}{16}\times10^{11}\times\:q^2}{0.1\times10^{-6}\times10}$

$\sf\implies\dfrac{1}{\sqrt{3}}=\dfrac{\frac{9}{16}\times10^{11}\times\:q^2}{0.1\times10^{-6}\times10}$

$\sf\implies\:q^2=\dfrac{16\times\sqrt{3}\times10^{-17}}{3\times9}$

$\sf\implies\:q=\sqrt{\dfrac{16\times\sqrt{3}\times10^{-17}}{3\times9}}$

$\sf\implies\:q\approx\:3.19\times10^{-9}C$

Therefore , The charge on each ball is approximately, $\sf3.19\times10^{-9}C$

Answer 2

$\Huge \fbox \red{an} \fbox \green{s} \fbox \pink{we} \fbox \purple{r}$

The charge on each ball is q^2 = 2.04 x 10^-17 C

Explanation:

We are given:

Mass of balls = 0.1 milligram

Length of threads = 0.4 m

Angle "θ" = 30°

Solution:

From equilibrium condition.

Tcos(30°) = mg

Tsin(θ) = F(E)

F(E) = K q^2 / 4.l^2 sin^2(30°) = K.q^2 / 2.l^2

Tanθ = F(E) / mg

θ = 30°

Tan(θ) = K.q^2 / 2 l^2 mg

2 l^2 mg / √ 3 . K = q^2

q^2 = 2 x (0.4)^2 (10^-7)(10) / √3 x 9 x 10^9

q^2 = 2 x 1.6 x 10^-7 / √3 x 9 x 10^9

q^2 = 3.2 x 10^-7 / 15.57 x 10^9

q^2 = 0.205 x 10^-7-9

q^2 = 2.05 x 10^-17 C

Thus the charge on each ball is q^2 = 2.04 x 10^-17 C

$\Huge \boxed{ \colorbox{yellow}{hope \: this \: helps}}$