Question

A small bob of mass 100 mg and charge +10 µC is connected to an insulating string of length 1 m. It is brought near to an infinitely long non-conducting sheet of charge density \( \sigma \) as shown in figure. If the string subtends an angle of 45° with the sheet at equilibrium, the charge density of sheet will be :

• A. 0.885 nC/cm\(^2\)
• B. 17.7 nC/cm\(^2\)
• C. 885 nC/cm\(^2\)
• D. 1.77 nC/cm\(^2\)

Hint:

When calculating the charge density from the angle, consider the forces acting on the object and how the electric field interacts with the charge. Use the equilibrium condition to set up the necessary equations.

Answer 1

The Correct Option is D

From the diagram in the solution, we have the force acting on the charge due to the electric field of the sheet: 
The force is given by: \[ F_e = qE = mg \] where \( q \) is the charge and \( E \) is the electric field due to the sheet. The electric field is related to the charge density \( \sigma \) as: \[ E = \frac{\sigma}{2 \epsilon_0} \] Thus, the equation becomes: \[ q \left( \frac{\sigma}{2 \epsilon_0} \right) = mg \] Rearranging to solve for \( \sigma \): \[ \sigma = \frac{2g m}{q} \] Substitute the known values: \[ \sigma = \frac{2 \times 8.85 \times 10^{-12} \times 100 \times 10^{-6} \times 10}{10 \times 10^{-6}} \] \[ \sigma = 17.7 \times 10^{-10} \text{ C/m}^2 \] \[ \sigma = 1.77 \text{ nC/cm}^2 \] Thus, the charge density of the sheet is \( 1.77 \) nC/cm\(^2\).

Answer 2

We need to determine the surface charge density \( \sigma \) of an infinite non-conducting sheet, given that a small bob of mass 100 mg and charge +10 µC, attached to a 1 m string, is in equilibrium when the string makes an angle of 45° with the sheet.

Concept Used:

The solution involves the principles of electrostatics and mechanics.

1. Electric Field of an Infinite Sheet: The electric field (\( E \)) produced by an infinite non-conducting sheet with a uniform surface charge density \( \sigma \) is constant and given by the formula: \[ E = \frac{\sigma}{2\epsilon_0} \] where \( \epsilon_0 \) is the permittivity of free space. The direction of the field is perpendicular to the sheet.
2. Electrostatic Force: The force (\( F_e \)) experienced by a charge \( q \) in an electric field \( E \) is: \[ F_e = qE \]
3. Equilibrium of Forces: For an object to be in static equilibrium, the net force acting on it must be zero. This means the vector sum of all forces is zero, which implies that the sum of the force components in any two perpendicular directions must also be zero. \[ \sum \vec{F} = 0 \quad \implies \quad \sum F_x = 0 \quad \text{and} \quad \sum F_y = 0 \]

 

Step-by-Step Solution:

Step 1: List the given values and convert them to SI units.

• Mass of the bob, \( m = 100 \, \text{mg} = 100 \times 10^{-6} \, \text{kg} \).
• Charge of the bob, \( q = +10 \, \mu\text{C} = +10 \times 10^{-6} \, \text{C} \).
• Angle subtended by the string with the vertical sheet is 45°. Since the sheet is vertical, this is also the angle the string makes with the vertical direction, \( \theta = 45^\circ \).
• We will use the acceleration due to gravity, \( g \approx 10 \, \text{m/s}^2 \), and the permittivity of free space, \( \epsilon_0 \approx 8.85 \times 10^{-12} \, \text{C}^2/\text{N}\cdot\text{m}^2 \).

Step 2: Analyze the forces acting on the bob in equilibrium.

The bob is subjected to three forces:

1. Gravitational Force (\( F_g \)): Acting vertically downwards. Its magnitude is \( F_g = mg \).
2. Electrostatic Force (\( F_e \)): The positively charged sheet repels the positive charge of the bob. This force acts horizontally, away from the sheet. Its magnitude is \( F_e = qE = q \frac{\sigma}{2\epsilon_0} \).
3. Tension (\( T \)): Acting along the string at an angle \( \theta = 45^\circ \) with the vertical.

Step 3: Apply the conditions for equilibrium by resolving the forces.

For the bob to be in equilibrium, the net force in both the horizontal and vertical directions must be zero. We resolve the tension \( T \) into its components:

• Horizontal component: \( T \sin\theta \)
• Vertical component: \( T \cos\theta \)

Balancing the forces:

In the horizontal direction:

\[ T \sin\theta = F_e \quad \cdots (1) \]

In the vertical direction:

\[ T \cos\theta = F_g \quad \cdots (2) \]

Step 4: Solve the equations to find an expression for the charge density \( \sigma \).

Divide equation (1) by equation (2):

\[ \frac{T \sin\theta}{T \cos\theta} = \frac{F_e}{F_g} \implies \tan\theta = \frac{F_e}{F_g} \]

Substitute the expressions for \( F_e \) and \( F_g \):

\[ \tan\theta = \frac{qE}{mg} = \frac{q(\frac{\sigma}{2\epsilon_0})}{mg} = \frac{q\sigma}{2mg\epsilon_0} \]

Now, rearrange the formula to solve for \( \sigma \):

\[ \sigma = \frac{2mg\epsilon_0 \tan\theta}{q} \]

Step 5: Substitute the numerical values to calculate \( \sigma \).

We have \( m = 100 \times 10^{-6} \text{ kg} \), \( g = 10 \text{ m/s}^2 \), \( \epsilon_0 = 8.85 \times 10^{-12} \text{ F/m} \), \( \theta = 45^\circ \) (so \( \tan 45^\circ = 1 \)), and \( q = 10 \times 10^{-6} \text{ C} \).

\[ \sigma = \frac{2 \times (100 \times 10^{-6} \, \text{kg}) \times (10 \, \text{m/s}^2) \times (8.85 \times 10^{-12} \, \text{C}^2/\text{N}\cdot\text{m}^2) \times (1)}{10 \times 10^{-6} \, \text{C}} \] \[ \sigma = \frac{2 \times 100 \times 10 \times 8.85 \times 10^{-12} \times 10^{-6}}{10 \times 10^{-6}} \] \[ \sigma = 2 \times 100 \times 8.85 \times 10^{-12} \] \[ \sigma = 1770 \times 10^{-12} \, \text{C/m}^2 = 1.77 \times 10^{-9} \, \text{C/m}^2 \]

The charge density of the sheet is \( 1.77 \times 10^{-9} \, \text{C/m}^2 \).