A point charge (m = 1.0 g) at the end of an insulating string of length L = 51 cm (Fig. 16-66) is observed to be in equilibrium in a uniform horizontal electric field of E = 9200 N/C, when the pendulum's position is as shown in Fig. 16-66, with the charge d = 1.0 cm above the lowest (vertical) position. If the field points to the right in Fig. 16-66, determine the magnitude and sign of the point charge.

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To determine the magnitude and sign of the point charge, we can use the equilibrium condition equation for an electric field and charge in a uniform electric field. The equation is:

F = q * E

Where:
F is the force on the charge
q is the charge magnitude
E is the electric field magnitude

In this case, the charge is in equilibrium, which means the net force on the charge is zero. The force due to gravity acting on the charge and the force due to the electric field balance each other out.

Let's analyze the forces acting on the charge:

1. Force due to gravity:
The force due to gravity can be calculated using the equation:

F_gravity = m * g

Where:
m is the mass of the charge
g is the acceleration due to gravity

In this case, the mass (m) is given as 1.0 g, which is equivalent to 0.001 kg. The acceleration due to gravity (g) is approximately 9.8 m/s^2.

F_gravity = (0.001 kg) * (9.8 m/s^2)

2. Force due to the electric field:
The force due to the electric field can be calculated using the equation:

F_electric = q * E

Where:
q is the charge magnitude
E is the electric field magnitude

We can rearrange the equation to solve for q:

q = F_electric / E

In this case, the electric field (E) is given as 9200 N/C. We'll use the value of F_electric found from the equilibrium condition to calculate q later.

Since the charge is in equilibrium, we can equate the force due to gravity and the force due to the electric field:

F_gravity = F_electric

Now we can substitute the expressions for each force:

(0.001 kg) * (9.8 m/s^2) = F_electric

Simplifying the equation:

0.0098 N = F_electric

Now we can substitute the calculated value of F_electric into the equation for q:

q = F_electric / E

q = (0.0098 N) / (9200 N/C)

Calculating the result:

q ≈ 1.065 × 10^-6 C

The magnitude of the point charge is approximately 1.065 × 10^-6 Coulombs. The sign of the charge can be determined based on the direction of the electric field. Since the field points to the right in the given diagram, the charge must be negative in order for the force due to the electric field to balance the force due to gravity. Therefore, the charge is negative.

To determine the magnitude and sign of the point charge, we can use the principle of equilibrium in both the gravitational and electric forces acting on the point charge.

Here are the steps to find the magnitude and sign of the point charge:

1. Identify the forces acting on the charge: In this case, we have the gravitational force due to the mass of the charge acting downward and the electric force due to the electric field acting horizontally.

2. Set up the equilibrium equations: Since the charge is in equilibrium, the forces must balance each other out. The sum of the vertical forces must be zero, and the sum of the horizontal forces must also be zero.

3. Equate the vertical forces: The only vertical force acting on the charge is the gravitational force. We can calculate it using the formula F_gravity = m * g, where m is the mass of the charge and g is the acceleration due to gravity (approximated as 9.8 m/s^2).

4. Equate the horizontal forces: The only horizontal force acting on the charge is the electric force. We can calculate it using the formula F_electric = q * E, where q is the magnitude of the charge and E is the electric field strength.

5. Solve for the charge magnitude and sign: Since the charge is in equilibrium, the electric force must balance the gravitational force. Therefore, we can equate F_gravity with F_electric and solve for q.

Here is the mathematical representation of the equilibrium equations:

Vertical equilibrium: F_gravity = 0
m * g = 0

Horizontal equilibrium: F_electric = 0
q * E = 0

Solving the equations, we find:

m * g = q * E
(1.0 g)(9.8 m/s^2) = q(9200 N/C)

Converting the mass to kg:
(0.001 kg)(9.8 m/s^2) = q(9200 N/C)

Simplifying the equation:

0.0098 N = q(9200 N/C)

Solving for q:

q = (0.0098 N) / (9200 N/C) ≈ 1.07 × 10^-6 C

Since the electric field points to the right (as shown in the figure), the point charge is positive. Thus, the magnitude and sign of the point charge are approximately 1.07 × 10^-6 C and +1.