Question

Two identical pith balls are charged by rubbing one against the other. They are suspended from a horizontal rod through two strings of length 20 cm each, the separation between the suspension points being 5 cm. In equilibrium, the separation between the balls is 3 cm. Find the mass of each ball and the tension in the strings. The charge on each ball has a magnitude 2.0 × 10−8 C.

Answer 1

Explanation:

Step 1:

Let the string tension be T, and let the force of attraction between the two balls be F.  

From the balls free-body diagram, we get

Tcosθ = mg    …eq^n (1)

Tsinθ = F    …eq^n (2)

Step 2:

From triangle ABC,

$\sin \theta=\frac{1}{20}$

$\cos \theta=\sqrt{1-\left(\frac{1}{20}\right)^{2}}$

Step 3:

divide equation (1) by  (2),

$\frac{\mathrm{T} \sin \theta}{\mathrm{T} \cos \theta}=\frac{F}{m g}$

$\tan \theta=\frac{F}{m g}$

$=\frac{F}{g \tan \theta}$

$m=\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2}}{r^{2}} \times\left(\frac{1}{g \tan \theta}\right)$

$m=\frac{9 \times 10^{9} \times\left(2.0 \times 10^{-8}\right)^{2}}{9 \times 10^{-4} \times 0.98 \times \tan \theta}$  

$m=\frac{9 \times 10^{9} \times 4 \times 10^{-16}}{8.82 \times 10^{-4} \times \tan \theta}$

$m=\frac{36 \times 10^{-7}}{8.82 \times 10^{-4} \times \frac{1}{2}}$

$m=\frac{36 \times 10^{-3}}{8.82 \times \frac{1}{2}}$  

$m=\frac{36 \times 10^{-3}}{4.41}$

$m=\frac{0.036}{4.41}$

$m=8.16 g$

$T=\frac{F}{\sin \theta}$

$=\frac{9 \times 10^{9} \times\left(2 \times 10^{-8}\right)^{2}}{9 \times 10^{-4} \times \sin \theta}$

$=\frac{10^{13} \times 4 \times 10^{-16}}{\sin \theta}$

$\sin \theta=\frac{1}{20}$     according to figure  

$=\frac{4 \times 10^{-3}}{\frac{2}{20}}$

$=80 \times 10^{-3}$

$=8 \times 10^{-2} N$

Answer 2

Given,

Two identical pith balls are charged by rubbing one against the other. They are suspended from a horizontal rod through two strings of length 20 cm each, the separation between the suspension points being 5 cm. In equilibrium, the separation between the balls is 3 cm.

To find:

Find the mass of each ball and the tension in the strings.

Solution:

The balls will develop opposite charges due to rubbing.

Please refer to the figure attached which is giving you free body diagram of ball at equilibrium position.

To gain equilibrium,

T sinθ = kq² / r² ...(i)

T cosθ = mg ...(ii)

Then, tanθ = kq²/r²mg

or, 1 / √[(20)² - 1²] = [ k.(2*10^-8)²] / [ (3*10^-2)²m*10 ]

or, m = 7.96 g

Substituing the value of m in eq.(ii):

T = mg / cosθ

or, T = 7.96*10^-3*10*20 / √[(20)² - 1²]

or, T = 7.72 * 10⁻² N

Therefore, Mass of each ball = 7.96 g

and, Tension is the string = 7.72 * 10⁻² N