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Let's break down the problem from the image.

We have a charged plate capacitor with plates placed vertically, separated by a distance of 6 cm (0.06 m). A lead ball with a mass of 0.8 g (0.0008 kg) is attached to a thread and carries a charge of \(2.1 \times 10^9 \, \text{C}\). The ball is deflected, forming an angle of 5° with the vertical due to the electrostatic force.

### Task Breakdown:

a) **Sketch the forces acting on the ball:**
   - The forces acting on the ball include:
     1. Gravitational force \( F_g = mg \) acting vertically downward.
     2. Tension \( T \) in the thread at an angle of 5° from the vertical.
     3. Electrostatic force \( F_e \) acting horizontally (perpendicular to the gravitational force), caused by the charged plates.

b) **Calculate the voltage at the capacitor plates:**
   - To calculate the voltage, we first find the electrostatic force \( F_e \) acting on the ball.
   - The electrostatic force can be determined from the angle of deflection, which will give us the electric field strength \( E \). From \( E \), we can find the voltage \( V \) using the relationship \( V = E \cdot d \), where \( d \) is the plate separation (0.06 m).

c) **Calculate the electric field strength inside the plate capacitor:**
   - The electric field strength \( E \) can be calculated from the balance of forces between the gravitational and electrostatic forces using trigonometry.

Let’s start by calculating the values for part (b) and part (c):

1. The forces are in equilibrium, so we can relate them using the angle:
   - \( F_e = T \sin(5^\circ) \) (horizontal component of tension),
   - \( F_g = T \cos(5^\circ) \) (vertical component of tension).

2. From \( F_g = mg \), calculate the gravitational force, and from this, deduce the electrostatic force.

Let me calculate these values for you.The gravitational force acting on the ball is \( F_g = 0.00785 \, \text{N} \), and the electrostatic force is \( F_e = 0.00069 \, \text{N} \).

Next, to calculate the electric field strength \( E \) inside the capacitor:

\[
F_e = q \cdot E \implies E = \frac{F_e}{q}
\]

Finally, to determine the voltage \( V \) between the plates, we can use:

\[
V = E \cdot d
\]

Let me calculate these now.The electric field strength inside the capacitor is approximately \( 326,958 \, \text{N/C} \), and the voltage between the plates is approximately \( 19,617 \, \text{V} \).

### Summary of Results:
- The electrostatic force on the ball is \( 0.00069 \, \text{N} \).
- The gravitational force is \( 0.00785 \, \text{N} \).
- The electric field strength inside the capacitor is \( 326,958 \, \text{N/C} \).
- The voltage across the capacitor plates is \( 19,617 \, \text{V} \).

Would you like further details on any part of the solution?

### Related Questions:
1. How does changing the mass of the ball affect the voltage required?
2. What would happen if the charge on the ball was doubled?
3. How would the electric field change if the plate separation increased?
4. What happens to the deflection angle if the voltage is reduced?
5. How do we derive the relationship between electrostatic force and electric field?

**Tip**: Always ensure units are consistent, especially when dealing with forces and charges in electric field problems.

Between the plates of a charged plate capacitor, which are placed vertically at a distance of 6 cm, there is a lead ball with a mass of 0.8 g attached to a thin insulating thread. It carries the charge 2.1 x 10^9 C.
a) Sketch the forces acting on the ball.
b) Calculate the voltage at the capacitor plates at which the ball is deflected so far that the thread forms an angle of 5° with the vertical.
c) In this case, calculate the electric field strength inside the plate capacitor.

Learn how to calculate the voltage and electric field in a charged plate capacitor problem with a lead ball deflected by electrostatic force. This problem involves electrostatics, forces in equilibrium, and trigonometry. Suitable for students in Grades 11-12.

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