Question

Two identical non-conducting solid spheres of same mass and charge are suspended in the air from a common point by two non-conducting, massless strings of the same length. At equilibrium, the angle between the strings is $\alpha$. The spheres are now immersed in a dielectric liquid of density $800kg{m}^{-3}$ and dielectric constant $21$. If the angle between the strings remains the same after the immersion, then

• A. electric force between the spheres remains unchanged
• B. the electric force between the spheres reduces
• C. the mass density of the spheres is $840kg{m}^{-3}$
• D. the tension in the strings holding the spheres remains unchanged

Answer 1

The correct option is C

the mass density of the spheres is $840kg{m}^{-3}$

Step 1: Given data

The angle between the strings is $\alpha$

The density of the liquid, ${\rho }_l=800kg{m}^{-3}$

Dielectric constant, $k=21$.

Here, $m$ is the mass of non-conducting solid spheres, $g$ is the acceleration due to gravity, ${\rho }_s$ is the mass density of the spheres, $q$ is the mass of the spheres, and ${\epsilon }_r$ is the relative permittivity of the medium.

Step 2: In the case of option B,

The force between two charges is given by Coulomb's law as,
$F=\frac{1}{4{\mathrm{\pi \epsilon }}_0{\mathrm{\epsilon }}_{\mathrm{r}}}·\frac{q_1{q}_2}{r^2}$
Where $q_1$ and $q_2$ are the two charges, $r$ is the distance between them, ${\epsilon }_0$ is the permittivity of free space and ${\epsilon }_r$ is the relative permittivity (also called the dielectric constant) of the medium.

The dielectric constant of air is $1$.
So the electric force between the charges in air is,
$F=\frac{1}{4{\mathrm{\pi \epsilon }}_0}·\frac{q_1{q}_2}{r^2}$

When the system is immersed in the liquid whose dielectric constant is $21$, the electric force between the charges is
$F^{\text{'}}=\frac{1}{4{\mathrm{\pi \epsilon }}_0·21}·\frac{q_1{q}_2}{r^2}$

Thus,
$F\text{'}=\frac{F}{21}...\left(1\right)$

Hence, option B is correct.

Step 3: In case of option C

From Newton's second law of motion, we have that,
$F=ma$
Where $m$ is the mass of the object and $a$ is the acceleration of the object when force $F$ is applied on the object.

At equilibrium, the net force along the y-axis is $0$

$T\cos \left(\frac{\alpha }{2}\right)=mg...\left(2\right)$
where $T_1$ is the tension in the string

When a liquid medium is introduced, the buoyant force acts opposite to gravity. So the above equation becomes,
$T^{\text{'}}\cos \left(\frac{\alpha }{2}\right)=mg-F_B$
Where $F_B$ is the buoyant force.

The buoyant force is given as,
$F_B=\rho gV$
Where $\rho$ is the density of the liquid, $V$ is the volume of the object the force is acting on.

Therefore,
$T^{\text{'}}\cos \left(\frac{\alpha }{2}\right)=mg-\rho gV...\left(3\right)$

Dividing $\left(3\right)$ by $\left(2\right)$,
$\begin{array}{ccc}& \frac{T^{\text{'}}\cos \left({\displaystyle \frac{\alpha }{2}}\right)}{T\cos \left({\displaystyle \frac{\alpha }{2}}\right)}=\frac{mg-{\rho }_{l}gV}{mg}& \\ \Rightarrow & \frac{T^{\text{'}}}{T}=1-\frac{{\rho }_l}{{\rho }_s}& \left[\because m=\rho V\right]...\left(4\right)\end{array}$

At equilibrium, the net force along the x-axis is $0$

Thus in air,
$T\cos \left(90-\frac{\alpha }{2}\right)=F$

In the liquid medium,
$T^{\text{'}}\cos \left(90-\frac{\alpha }{2}\right)=F^{\text{'}}$

But,
$F^{\text{'}}=\frac{F}{21}$

So,
$\begin{array}{cc}& T^{\text{'}}\cos \left(90-\frac{\alpha }{2}\right)=\frac{F}{21}\\ \Rightarrow & T^{\text{'}}\cos \left(90-\frac{\alpha }{2}\right)=\frac{T\cos \left(90-{\displaystyle \frac{\alpha }{2}}\right)}{21}\\ \Rightarrow & T^{\text{'}}=\frac{T}{21}\\ \Rightarrow & \frac{T^{\text{'}}}{T}=\frac{1}{21}...\left(5\right)\end{array}$

Substituting $\left(5\right)$ in $\left(4\right)$, we have,
$\begin{array}{cc}\Rightarrow & \frac{1}{21}=1-\frac{{\rho }_l}{{\rho }_s}\\ \Rightarrow & \frac{800}{{\rho }_s}=\frac{20}{21}\\ \Rightarrow & {\rho }_s=840kg{m}^{-3}\end{array}$

Hence, option C is also correct.

Therefore, only options B and C are correct.