Suspended Charged Spheres - Detailed Solution & Interactive Quiz

Charged Spheres Equilibrium Problem [GUJCET 2014]

Question: Two spheres having the same radius and mass are suspended by two strings of equal length from the same point in such a way that their surfaces touch each other. When a total charge of \(4 \times 10^{-6}\, \text{C}\) is deposited on them, the charge divides equally so that each sphere gets \(2 \times 10^{-6}\, \text{C}\). They repel each other and, in equilibrium, the angle between their strings becomes \(60^\circ\). Given that the distance from the point of suspension to the center of a sphere is \(10\,\text{cm}\), find the mass of each sphere. (Take \( k = 9 \times 10^9\, \text{N·m}^2/\text{C}^2 \) and \( g = 10\, \text{m/s}^2\).)

(a) \(0.3117\, \text{kg}\)

(b) \(0.6235\, \text{kg}\)

(c) \(0.1559\, \text{kg}\)

(d) \(1.2468\, \text{kg}\)

Detailed Step-by-Step Explanation

Step 1: Setup of the Problem
Two identical spheres of mass \(m\) are suspended by strings of length \(L = 10\, \text{cm} = 0.1\, \text{m}\). In equilibrium, the angle between the strings is \(60^\circ\). This means each string makes an angle of \(30^\circ\) with the vertical.

Step 2: Force Analysis
For each sphere, the forces acting are:
- Weight: \( mg \) (downward)
- Tension in the string: \( T \) (along the string)
- Electrostatic repulsive force: \( F_e \) (horizontal)

Resolving the tension into vertical and horizontal components:

Tension vertical component: \( T\cos30^\circ = mg \)   (1)
Tension horizontal component: \( T\sin30^\circ = F_e \)  (2)

Dividing (2) by (1):

\( \frac{T\sin30^\circ}{T\cos30^\circ} = \frac{F_e}{mg} \quad \Rightarrow \quad \tan30^\circ = \frac{F_e}{mg} \)

Since \(\tan30^\circ = \frac{1}{\sqrt{3}}\), we have:

\( F_e = \frac{mg}{\sqrt{3}} \)   (3)

Step 3: Electrostatic Force
Each sphere carries a charge \( q = \frac{4 \times 10^{-6}}{2} = 2 \times 10^{-6}\, \text{C} \). The electrostatic repulsive force between two point charges separated by a distance \( d \) is given by:

\( F_e = \frac{kq^2}{d^2} \)

Step 4: Distance Between the Centers
Each sphere is displaced horizontally by:

\( L\sin30^\circ = 0.1 \times 0.5 = 0.05\, \text{m} \)

Hence, the distance between the centers is:

\( d = 2 \times 0.05 = 0.1\, \text{m} \)

Step 5: Equating Forces
From (3) and the expression for \( F_e \):

\( \frac{mg}{\sqrt{3}} = \frac{kq^2}{d^2} \)

Substitute \( d = 0.1\, \text{m} \), \( k = 9 \times 10^9\, \text{N·m}^2/\text{C}^2 \), and \( g = 10\, \text{m/s}^2 \):

\( \frac{mg}{\sqrt{3}} = \frac{9 \times 10^9 \times (2 \times 10^{-6})^2}{(0.1)^2} \)

Step 6: Solve for \(m\)
First, calculate \( q^2 \):
\( (2 \times 10^{-6})^2 = 4 \times 10^{-12} \)

And \((0.1)^2 = 0.01\). Thus:

\( \frac{mg}{\sqrt{3}} = \frac{9 \times 10^9 \times 4 \times 10^{-12}}{0.01} \)

Simplify the right-hand side:
\( 9 \times 10^9 \times 4 \times 10^{-12} = 36 \times 10^{-3} = 0.036 \)
Dividing by \(0.01\):
\( \frac{0.036}{0.01} = 3.6 \)

So we have:

\( \frac{mg}{\sqrt{3}} = 3.6 \)

Solve for \(m\):

\( m = \frac{3.6 \sqrt{3}}{g} \)
Substitute \( g = 10 \):
\( m = \frac{3.6 \sqrt{3}}{10} \)

Now, using \(\sqrt{3} \approx 1.732\):

\( m \approx \frac{3.6 \times 1.732}{10} = \frac{6.2352}{10} = 0.62352\, \text{kg} \)

Final Answer: The mass of each sphere is approximately \(0.6235\, \text{kg}\), which corresponds to option (b).