Question

Two small spheres each of mass $10mg$ are suspended from a point by threads $0.5m$ long. They are equally charged and repel each other to a distance of $0.20m$. Then charge on each of the sphere is $\left(\frac{a}{21}\right)\times {10}^{-8}C$. The value of ‘a’ will be ________. [Take $g=10m/s^2$]

Answer 1

Step 1: Given

Mass of each sphere: $m=10mg=10\times {10}^{-6}kg$

Length of threads: $L=0.5m$

Distance between masses: $d=0.20m$

Charge on each sphere: $q_1=q_2=q=\left(\frac{a}{21}\right)\times {10}^{-8}C$

Acceleration due to gravity: $g=10m/s$

Step 2: Formula Used

$F=k\frac{q_1{q}_2}{d^2}$

Step 3: Find an expression for $\tan \theta$

Find the value of $T\cos \theta$ which will be equal to the weight of the mass (acting downwards) in order to balance it.

$\begin{array}{rcl}T\cos \theta & =& w\\ & =& mg\\ & =& 10\times {10}^{-6}\times 10\\ & =& {10}^{-4}\dots \left(1\right)\end{array}$

Find the value of $T\sin \theta$ which will be equal to the electrostatic force $F_e$ acting on the charge in order to balance it.

$\begin{array}{rcl}T\sin \theta & =& F_e\\ & =& k\frac{q_1{q}_2}{d^2}\\ & =& k\frac{q^2}{d^2}\\ & =& 9\times {10}^9\frac{q^2}{0.2^2}\dots \left(2\right)\end{array}$

Find an expression for $\tan \theta$ by dividing equation 2 by equation 1

$\begin{array}{rcl}\tan \theta & =& \frac{T\sin \theta }{T\cos \theta }\\ & =& \frac{9\times {10}^9\frac{q^2}{0.2^2}}{{10}^{-4}}\\ & =& \frac{9\times {10}^9\frac{q^2}{0.2^2}}{{10}^{-4}}\\ & =& 2.25\times {10}^{15}{q}^2\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \dots \end{array}\begin{array}{rcl}& & \left(3\right)\end{array}\\ & & \end{array}$

Find an expression for $\tan \theta$ using trigonometry

$\begin{array}{rcl}\tan \theta & =& \frac{Perpendicular}{Base}\\ & =& \frac{\frac{d}{2}}{L}\\ & =& \frac{0.20}{2\times 0.5}\\ & =& 0.20\cdots \left(4\right)\end{array}$

Step 4: Find the value of a

Find the value of charge by equating both the values of tangent of angle, that is, equate equation 3 and equation 4.

$\begin{array}{rcl}2.25\times {10}^{15}{q}^2& =& 0.20\\ & \Rightarrow & q^2=\frac{0.20}{2.25\times {10}^{15}}\\ & \Rightarrow & q=\sqrt{\frac{0.20}{2.25\times {10}^{15}}}\\ & \Rightarrow & q=\sqrt{\frac{200}{225}\times {10}^{-16}}\end{array}$

Substitute the value of charge obtained in $q=\left(\frac{a}{21}\right)\times {10}^{-8}C$.

$$\begin{gathered} \left(\frac{a}{21}\right)\times {10}^{-8}=\begin{array}{l}\sqrt{\frac{200}{225}\times {10}^{-16}}\end{array} \\ \Rightarrow \left(\frac{a}{21}\right)\times {10}^{-8}=\begin{array}{l}\sqrt{\frac{8}{9}}\times {10}^{-8}\end{array} \\ \Rightarrow a=\frac{21\sqrt{8}}{3} \\ \Rightarrow a=19.799C \end{gathered}$$

Hence, Two small spheres each of mass $10mg$ are suspended from a point by threads $0.5m$ long. They are equally charged and repel each other to a distance of $0.20m$. Then charge on each of the sphere is $\left(\frac{a}{21}\right)\times {10}^{-8}C$. The value of ‘a’ will be $\approx 20C$