Question

Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle of ${30}^{\circ }$ with each other. When suspended in a liquid of density $0.8gc{m}^{-3}$, the angle remains the same. If the density of the material of the sphere is $1.6gc{m}^{-3}$, the dielectric constant of the liquid is
A). 4
B). 3
C). 2
D). 1

Answer 1

Hint In order to solve this question first we have to draw a proper diagram and then we will proceed further by equating the forces applied on the ball with the help of the diagram. We will use the concept of trigonometry along with the concept of electrostatics.
Formula used: $w=mg,\text{Upthrust = }V\sigma g,V={\displaystyle \frac{m}{\rho }}$ where m= mass of the ball , g =acceleration due to gravity , $V$ = volume of the submerged portion, $\sigma$= density of the liquid , $\rho$= density of the ball

Complete step-by-step solution:
Figure:

Initially, the forces acting on each ball are
(i) Tension T
(ii) Weight mg (m=mass of the ball in kg , g = acceleration due to gravity in $m{s}^{-2}$)
(iii) The electrostatic force of repulsion F for its equilibrium along vertical, where $\theta$ is the angle made by the string and the vertical
$T\cos \theta =mg\to (1)$
And along horizontal, we have
$T\sin \theta =F\to (2)$
 Dividing equation (2) by (1), we get
$$\begin{gathered} {\displaystyle \frac{T\sin \theta }{T\cos \theta }}={\displaystyle \frac{F}{mg}} \\ \Rightarrow {\displaystyle \frac{\sin \theta }{\cos \theta }}={\displaystyle \frac{F}{mg}} \\ \Rightarrow \tan \theta ={\displaystyle \frac{F}{mg}}\to (3) \end{gathered}$$
Now ,when the balls are suspended in a liquid of density σ and dielectric constant $K$ , the electrostatic force will become ${\displaystyle \frac{1}{K}}$ times, i.e.$F^{\prime }={\displaystyle \frac{F}{K}}$
 while weight becomes original weight - Upthrust.
But before that let us know about Upthrust, which can be defined as the upward force applied by a liquid on an object immersed (partially or fully) or floating in the liquid. Mathematically it can be written as $\text{Upthrust = }V\sigma g$. This is also known as the buoyant force.
$$\begin{gathered} m{g}^{\prime }=mg-\text{Upthrust} \\ \Rightarrow m{g}^{\prime }=mg-V\sigma g \\ \Rightarrow m{g}^{\prime }=mg-\left({\displaystyle \frac{m}{\rho }}\right)\sigma g\left[\because V={\displaystyle \frac{m}{\rho }}\right] \end{gathered}$$
Taking mg as common , we get
$\Rightarrow m{g}^{\prime }=mg\left[1-{\displaystyle \frac{\sigma }{\rho }}\right]\to (4)$
Now ,for the equilibrium of the balls we need a relation between the new or apparent force and the apparent weight, given as below
$F^{\prime }=m{g}^{\prime }\tan {\theta }^{\prime }\to (5)$
Angle changes from$\theta \text{ to }\theta \text{'}$ because of the apparent force applied
Now , Let us substitute the value of mg’ from equation (4) in equation (5)
So we get:
$F^{\prime }=mg\left[1-{\displaystyle \frac{\sigma }{\rho }}\right]\tan {\theta }^{\prime }$
Let us find the value of $\tan {\theta }^{\prime }$ from the above equation:
$\tan {\theta }^{\prime }={\displaystyle \frac{F^{\prime }}{mg\left[1-{\displaystyle \frac{\sigma }{\rho }}\right]}}$
And we know $F^{\prime }={\displaystyle \frac{F}{K}}$
$$\begin{gathered} \tan {\theta }^{\prime }={\displaystyle \frac{{\displaystyle \frac{F}{K}}}{mg\left[1-{\displaystyle \frac{\sigma }{\rho }}\right]}} \\ \Rightarrow \tan {\theta }^{\prime }={\displaystyle \frac{F}{Kmg\left[1-{\displaystyle \frac{\sigma }{\rho }}\right]}}\to (6) \end{gathered}$$
As we know from the given problem statement:
$$\begin{gathered} \theta ={\theta }^{\prime } \\ \Rightarrow \tan \theta =\tan {\theta }^{\prime } \end{gathered}$$
Now, from equation (6) and equation (3), we have
$$\begin{gathered} \tan \theta =\tan {\theta }^{\prime } \\ {\displaystyle \frac{F}{mg}}={\displaystyle \frac{F}{Kmg\left[1-{\displaystyle \frac{\sigma }{\rho }}\right]}} \end{gathered}$$
Let us now solve this equation to find the value of the dielectric constant K.
$$\begin{gathered} {\displaystyle \frac{F}{mg}}={\displaystyle \frac{F}{Kmg\left[1-{\displaystyle \frac{\sigma }{\rho }}\right]}} \\ \Rightarrow K={\displaystyle \frac{mg}{mg\left[1-{\displaystyle \frac{\sigma }{\rho }}\right]}} \\ \Rightarrow K={\displaystyle \frac{1}{\left[1-{\displaystyle \frac{\sigma }{\rho }}\right]}} \end{gathered}$$
As we have been given in the problem, the density is $0.8gc{m}^{-3}$ and the density of the material of the sphere is $1.6gc{m}^{-3}$. So, let us substitute the values in the above equation to find the value of dielectric constant K.
$$\begin{gathered} K={\displaystyle \frac{1}{\left[1-{\displaystyle \frac{\sigma }{\rho }}\right]}} \\ \Rightarrow K={\displaystyle \frac{1}{\left[1-{\displaystyle \frac{0.8}{1.6}}\right]}} \\ \Rightarrow K={\displaystyle \frac{1}{\left[1-{\displaystyle \frac{1}{2}}\right]}} \\ \Rightarrow K={\displaystyle \frac{1}{\left[{\displaystyle \frac{2-1}{2}}\right]}} \\ \Rightarrow K={\displaystyle \frac{1\times 2}{1}} \\ \Rightarrow K=2 \end{gathered}$$
Hence, the dielectric constant of the liquid is 2
So, the correct answer is option C.

Note: The relative permittivity of a substance, or dielectric constant, is its permittivity expressed as a ratio to the permittivity of the vacuum. Permittivity is a material property that has an effect on the Coulomb force in the material between two point charges. Two or more solid objects in touch are exerting forces on one another. We give different names to those contact forces based on the objects in contact. When a loop, cord, wire, or spring happens to be one of those objects in contact, we call friction as stress.