A parallel-plate capacitor is constructed of two horizontal 15.2-cm-diameter circular plates. A 1.9 g plastic bead, with a charge of -5.6 nC is suspended between the two plates by the force of the electric field between them.
What is the charge on the positive plate?
To find the charge on the positive plate of the parallel-plate capacitor, we need to first find the electric field between the plates caused by the charge on the plastic bead. Then we can use the concept of capacitance to determine the charge on the plate.
1. Find the electric field between the plates:
- The electric field between the plates of a parallel-plate capacitor is given by the equation:
E = V/d
where E is the electric field, V is the voltage across the plates, and d is the distance between the plates.
- In this case, the information about the diameter of the plates is not directly useful for finding the electric field. However, we can find the distance between the plates using the diameter of the plates.
d = diameter/2
d = 15.2 cm / 2
- Convert the distance to meters:
d = 15.2 cm / 2 * (1 m / 100 cm)
- Now, we need to find the voltage across the plates. Since only the plastic bead is mentioned, assume there is no other charge on or between the plates. In this case, the voltage across the plates is zero, since the electric potential difference (voltage) is the difference in electric potential energy per unit charge between two points.
V = 0
- Substitute the values into the equation to find the electric field:
E = 0 / (15.2 cm / 2 * (1 m / 100 cm))
2. Calculate the electric force on the plastic bead:
- The electric force on a charged particle in an electric field is given by the equation:
F = qE
where F is the electric force, q is the charge of the particle, and E is the electric field.
- Substitute the given charge and the calculated electric field into the equation:
F = (-5.6 nC) * E
3. Calculate the weight of the plastic bead:
- Weight is the force due to gravity acting on an object with mass m. The weight is given by the equation:
W = mg
where W is the weight, m is the mass, and g is the acceleration due to gravity.
- The mass of the plastic bead is given as 1.9 g:
m = 1.9 g
- The acceleration due to gravity is approximately 9.8 m/s^2:
g = 9.8 m/s^2
- Substitute the given mass and acceleration due to gravity into the equation:
W = (1.9 g) * (9.8 m/s^2)
4. Equate the electric force on the plastic bead to its weight:
- Since the plastic bead is suspended between the plates, the electric force on it will balance its weight.
F = W
- Equate the two forces:
(-5.6 nC) * E = (1.9 g) * (9.8 m/s^2)
5. Calculate the electric field:
- Rearrange the equation to solve for the electric field:
E = ((1.9 g) * (9.8 m/s^2)) / (-5.6 nC)
Now that we have the value of the electric field, let's calculate the charge on the positive plate using the concept of capacitance.
6. Calculate the charge on the positive plate:
- The capacitance of a parallel-plate capacitor is given by the equation:
C = ε₀ * (A / d)
where C is the capacitance, ε₀ is the permittivity of free space, A is the area of each plate, and d is the distance between the plates.
- The permittivity of free space is approximately 8.854 x 10^-12 F/m.
- The area of each plate can be calculated using the formula for the area of a circle: A = π * (radius^2)
- The radius can be found using the diameter: radius = diameter/2
- Substitute the values of the permittivity of free space, radius, and distance into the equation to find the capacitance:
C = (8.854 x 10^-12 F/m) * (π * ((15.2 cm / 2 * (1 m / 100 cm))^2))
- The charge on a capacitor is given by the equation:
Q = C * V
where Q is the charge on the capacitor, C is the capacitance, and V is the voltage across the plates.
- In this case, the voltage across the plates is zero, as mentioned earlier.
V = 0
- Substitute the capacitance and voltage into the equation to find the charge on the positive plate:
Q = (8.854 x 10^-12 F/m) * (π * ((15.2 cm / 2 * (1 m / 100 cm))^2)) * 0
The result of the calculation will be zero, indicating that the charge on the positive plate of the parallel-plate capacitor is zero.