Two parallel conducting plates are separated by 10.0 cm, and one of them is taken to be at zero volts.
(a) What is the electric field strength between them, if the potential 8.50 cm from the zero volt plate (and 1.50 cm from the other) is 540 V?
(b) What is the voltage between the plates?
To find the electric field strength between the parallel conducting plates, we can use the formula:
Electric field strength (E) = Voltage (V) / Distance (d)
(a) To find the electric field strength at a given point between the plates, let's use the given information. We are given that the potential 8.50 cm from the zero volt plate (and 1.50 cm from the other) is 540 V.
Given:
Voltage (V) = 540 V
Distance (d) = 8.50 cm = 0.085 m
Now, we can plug these values into the formula to find the electric field strength (E) between the plates:
E = V / d
E = 540 V / 0.085 m
E ≈ 6352.94 V/m
Therefore, the electric field strength between the plates is approximately 6352.94 V/m.
(b) To find the voltage between the plates, we can rearrange the formula:
Voltage (V) = Electric field strength (E) * Distance (d)
Given:
Electric field strength (E) = 6352.94 V/m
Distance (d) = 10.0 cm = 0.1 m
Now, we can plug these values into the formula to find the voltage (V) between the plates:
V = E * d
V = 6352.94 V/m * 0.1 m
V ≈ 635.29 V
Therefore, the voltage between the plates is approximately 635.29 V.