Question

Two identical balls, each having a charge of $2.00\times {10}^{-7}C$ and a mass of $100g,$ are suspended from a common point by two insulating strings each 50 cm long. The balls are held at a separation 5.0 cm apart and then released. Find (a) the electric force on one of the charged balls (b) the components of the resultant force on it along and perpendicular to the string (c) the tension in the string (d) the acceleration of one of the balls. Answers are to be obtained only for the instant just after the release.

• A. $F=0.156Na=98m/se{c}^2$
• B. $F=0.196Na=53m/se{c}^2$
• C. $F=0.144Na=95m/se{c}^2$
• D. $F=0.199Na=86m/se{c}^2$

Answer 1

The correct option is C $F=0.144Na=95m/se{c}^2$

Two identical balls, each having a charge $=2.00\times {10}^{-7}C$

mass $\left(m\right)=100g$

string each $=50cm(50\times {10}^{-2}m)$

The balls are held at a separation apart $=5.0cm$ $\left(d\right)$

Find $-\left(a\right)$ electric force on one of the charged balls.

$\left(b\right)$ the components of the resultant force on it along and perpendicular to the string

$q_1=q_2=2\times {10}^{-7}C$

$m_1=m_2=100g=0.1kg$

$L=50cm=0.5m$

$d=5cm=5\times {10}^{-2}m$

The force of gravity is -

$mg=0.1\times 9.8=0.98N$

$\left(d\right)$ The electrostatic force between the two charges is -

$F=\frac{1}{4\pi {ϵ}_0}\frac{q^2}{d^2}$

$=9\times {10}^9\times \frac{{(2\times {10}^{-7})}^2}{{(5\times {10}^{-2})}^2}=0.144N$

$\left(a\right)$ from the geometry of the figure, we have

$\sin \theta =\frac{d}{2L}=\frac{5\times {10}^{-2}}{2\times 0.5}=0.05$

$\theta ={\sin }^{-1}\left(0.05\right)={2.87}^0$

The force of gravity and the electrostatic force between the two will have a resultant force $R$ as shown in the fig.

From the figure, we can that the resultant force has a component along the string and a component perpendicular to the string.

Hence, we have the component along the string as

$R_a=mg\cos \theta +F\cos (90-\theta )$

$R_a=0.98\times \cos 2.87+0.144+\cos ({90}^0-\theta )$

$=0.98\times \cos 2.87+0.144\times \cos 87.13$

$R_a=0.979+0.0072=0.986N$

Also, we have the component perpendicular the string as

$R_p=F\sin ({90}^0-\theta )-mg\sin \theta$

$R_p=0.144\times \sin 87.13-0.98\times \sin 2.87$

$R_p=0.1438-0.049=0.095N$

$\left(b\right)$ The radial component (component along the string) provides tension in the string hence, we have

$T=R_a=0.986N$

$\left(c\right)$ The perpendicular components provide the acceleration to the balls. Hence, we have

$a=\frac{R_p}{m}=\frac{0.095}{0.1}=0.95m/s^2$