Two identical non-conducting solid spheres of same mass and charge are suspended in air from a commo

Q: Why does the electric force decrease in the liquid?

The dielectric medium polarizes and creates an opposing electric field, reducing the net force between the charges by a factor of its dielectric constant.

Q: Why did we use effective weight instead of just subtracting buoyancy?

It's mathematically cleaner. $W_{eff} = mg - F_b = V\rho_s g - V\rho_l g = mg(1 - \rho_l/\rho_s)$. This allows us to easily cancel out the $mg$ term later.

Q: Does the tension in the string remain the same?

No! Both the horizontal and vertical forces have changed, so the resultant force (which tension balances) must also change.

Q: What if the angle had increased?

That would mean the reduction in effective weight (due to buoyancy) was more significant than the reduction in electric force (due to the dielectric).

Q: Why did we only analyze one sphere?

Due to symmetry, both spheres experience identical forces and angles. Analyzing one is completely sufficient to solve the system.

Question

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

Options:

(A) electric force between the spheres remains unchanged

(B) electric force between the spheres reduces

(C) mass density of the spheres is $840	ext{ kg m}^{-3}$

(D) the tension in the strings holding the spheres remains unchanged

Sigma Percentile · Accepted Answer

### The Setup in Air

Imagine two identical charged spheres suspended from a common point. They repel each other due to electrostatic forces, pushing apart until they reach a state of equilibrium.

At this equilibrium position, the strings make an angle $\alpha$ with each other. Let's focus on just one of the spheres and analyze the forces acting on it.

There are three primary forces at play here. First, gravity pulls the sphere downwards with a force $mg$. Second, the electrostatic repulsion $F_e$ pushes it horizontally away from the other sphere. Finally, the tension $T$ in the string pulls it along the length of the string.

To make sense of this, we resolve the tension $T$ into horizontal and vertical components. The vertical component $T \cos(\alpha/2)$ perfectly balances the weight $mg$. The horizontal component $T \sin(\alpha/2)$ balances the repulsive electric force $F_e$.

Dividing these two equations gives us our master relationship:
$$ 	an(\alpha/2) = \frac{F_e}{mg} $$

### The Dive into the Dielectric

Now, the entire system is submerged into a dielectric liquid. This changes the environment drastically.

First, the liquid exerts an upward buoyant force $F_b$ on the spheres. This effectively reduces the weight of the spheres. We can express this new effective weight as $W' = mg - F_b$.

Using the concept of density, the buoyant force is the weight of the displaced liquid. So, $W' = mg \left(1 - \frac{ho_l}{ho_s}ight)$, where $ho_l$ is the liquid's density and $ho_s$ is the sphere's density.

Second, the electrostatic force is weakened because the medium is no longer a vacuum. The new electric force $F_e'$ becomes the original force divided by the dielectric constant $\epsilon_r$. So, $F_e' = \frac{F_e}{\epsilon_r}$.

### The Unchanging Angle

The problem states a crucial condition: the angle $\alpha$ remains exactly the same after immersion. This is the key to unlocking the solution.

Since the spheres are again in equilibrium, our master equation still holds true with the new forces. We can write:
$$ 	an(\alpha/2) = \frac{F_e'}{W'} $$

Because the angle hasn't changed, the tangent of the angle must also be identical in both scenarios. We can equate the two expressions we derived.

$$ \frac{F_e}{mg} = \frac{F_e'}{W'} $$

### The Elegant Cancellation

Now, we substitute the expressions for the new force and the new effective weight into our equated tangents.

$$ \frac{F_e}{mg} = \frac{F_e / \epsilon_r}{mg \left(1 - \frac{ho_l}{ho_s}ight)} $$

Notice how beautifully the original electric force $F_e$ and the original weight $mg$ cancel out from both sides of the equation. We are left with a pure, dimensionless relationship.

$$ 1 = \frac{1}{\epsilon_r \left(1 - \frac{ho_l}{ho_s}ight)} $$

### Final Calculation

We can rearrange this elegant equation to solve for the unknown density of the sphere.

$$ 1 - \frac{ho_l}{ho_s} = \frac{1}{\epsilon_r} $$

$$ \frac{ho_l}{ho_s} = 1 - \frac{1}{\epsilon_r} $$

Now, we simply plug in the values provided in the problem. The density of the liquid $ho_l$ is $800 	ext{ kg/m}^3$, and the dielectric constant $\epsilon_r$ is $21$.

$$ \frac{800}{ho_s} = 1 - \frac{1}{21} $$

$$ \frac{800}{ho_s} = \frac{20}{21} $$

Solving for $ho_s$, we get:
$$ ho_s = \frac{800 	imes 21}{20} = 840 	ext{ kg/m}^3 $$

**Conclusion:** The mass density of the spheres is indeed $840 	ext{ kg/m}^3$, making option (C) correct. Furthermore, because the spheres are in a dielectric medium, the net electric force between them reduces by a factor of $21$, making option (B) correct as well.

Select Answer:

Visualized Solution

Equilibrium in Air

Force Balance

Master Equation

Immersed in Liquid

New Forces

New Equilibrium

Equating Tangents

Substitution

Simplification

Rearranging for Density

Plugging Values

Calculation

Final Answer

The Way Forward

The Sigma Insight: Coulomb's Law

The Setup in Air

The Dive into the Dielectric

The Unchanging Angle

The Elegant Cancellation

Final Calculation

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