Question

Two identical balls each having a density "ρ" are suspended from a common point by two insulating strings of equal length. Both the balls have equal mass and charge. In equilibrium each sting makes an angle "θ" with vertical. Now, both the balls are immersed in a liquid. As a result the angle "θ" does not change the density of liquid is "σ" Find the dielectric constant (K) of the liquid?

Ans:- $\sf{K = \dfrac{\rho}{\rho - \sigma}}$

Class:- 12th ; Physics
Chapter:- Electrostatics. ​

Answer 1

Answer:

In Air:

Let Electrostatic Force between the charges be F

In the 1st case , we can divide the Tension in the string as follows :

$1) \: \: T \sin( \theta) = F$

$2) \: \: T \cos( \theta) = mg$

Dividing the 2 equations, we get :

$\boxed{ \bold{ \tan( \theta) = \dfrac{F}{mg} }}$

Inside the liquid :

The Electrostatics force will be F/k considering the dielectric constant to be k.

Similarly the net weight of the blocks will be reduced due to Buoyant Force such as :

$= mg - F_{B}$

$= mg -( v \times \sigma \times g)$

$= mg - ( \dfrac{m}{ \rho} \times \sigma \times g)$

$= mg(1 - \dfrac{ \sigma}{ \rho} )$

As per the Question , the angle of the strings doesn't change :

So we can say that :

$\tan( \theta) = \dfrac{ \bigg( \dfrac{F}{k} \bigg) }{mg(1 - \dfrac{ \sigma}{ \rho}) }$

Equating the tan function in both the cases (i.e. of air and in liquid) :

$\therefore \: \: \dfrac{F}{mg} = \dfrac{ \bigg( \dfrac{F}{k} \bigg) }{mg(1 - \dfrac{ \sigma}{ \rho}) }$

$= > k = \dfrac{1}{ \bigg \{ \dfrac{ \rho - \sigma}{ \rho} \bigg \}}$

$= > k = \dfrac{ \rho}{ \rho - \sigma}$

$\boxed{ \red{ \sf{Hence \: \: Proved}}}$

Answer 2

The dielectric constant of the liquid :

$\dfrac{\rho}{\rho- \sigma}$

Explanation :

Given :

• Two identical balls each having a density "ρ"
• Balls are suspended from a common point by two insulating strings of equal length.
• The balls have equal mass and charge
• In equilibrium each sting makes an angle "θ" with vertical
• Both the balls are immersed in a liquid.
• The result angle "θ" does not change the density of liquid is "σ"

To Find :

The dielectric constant (K) of the liquid.

Solution :

According to the attachment ( In the Liquid ) :

There is Fe because it is given that the balls have same charge. We know that the same charge never attract each other. So, they will repel. So, there will be force due to electric.

There tension will also react. In attachment. It is denoted to "T", Near tension the angle will be theta as the first angle.

It is given that :

In equilibrium each sting makes an angle "θ" with vertical,

Hence,

After in liquid,

$\sf \implies Fe^{1} = \dfrac{Fe}{K}$

The thrust will be :

$\implies \sf W^1 = W - Upthrust$

We know the Lami's theorem :

In which if there are three forces,

Let the forces be A,B & C.

If the α make 1 angle, β make one angle & γ make one angle.

So, It would be :

$\implies \sf \dfrac{A}{sin\;\alpha}=\dfrac{B}{sin\;\beta} = \dfrac{C}{sin\;\gamma}$

Hence,

After applying Lami's Theorem,

It will be,

$\implies \sf \dfrac{W}{sin(90+\theta)}=\dfrac{Fe}{sin(180-\theta)}$

After solving,

$\implies \sf \dfrac{W}{cos\;\theta}=\dfrac{Fe}{sin\;\theta}\qquad\bigg[\bf Equation\;1\bigg]$

$\implies \sf \dfrac{W^1}{cos\;\theta} = \dfrac{Fe}{sin\;\theta}\qquad\bigg[\bf Equation\;2\bigg]$

_________________[ In Liquid ]

• Dividing Equation 1 with Equation 2,

$\implies \sf \dfrac{W}{W^1}=\dfrac{Fe}{Fe^1}\\ \\\\\implies \sf K= \dfrac{W}{W-\;Upthrust}\\ \\ \\ \\\implies \sf K = \dfrac{v\;\rho\;g}{v\rho\;g\;-v\sigma\;g}\qquad\bigg[\bf \dfrac{Fe} {Fe^1}= K\bigg]\\ \\ \\\implies \sf K = \dfrac{\rho}{\rho- \sigma}$