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 450 V?
kV/m
(b) What is the voltage between the plates?
V
To find the electric field strength between the parallel conducting plates and the voltage between the plates, we need to use the equation for electric field and potential difference.
(a) Electric Field Strength (E):
The electric field between the plates can be found using the formula:
E = V / d
Where:
E is the electric field strength,
V is the potential difference between the plates,
d is the distance between the plates.
In this case, the potential at a distance of 8.50 cm from the zero volt plate (1.50 cm from the other plate) is given as 450 V, and the distance between the plates (d) is 10.0 cm.
Therefore, the electric field strength (E) can be calculated as follows:
E = V / d
E = 450 V / 10.0 cm
First, let's convert the distance to meters:
10.0 cm = 10.0 cm * (1 m / 100 cm) = 0.10 m
Now, we can calculate the electric field strength:
E = 450 V / 0.10 m
E = 4500 V/m
Therefore, the electric field strength between the plates is 4500 V/m or 4.5 kV/m.
(b) Voltage between the Plates:
The voltage between the plates can be found using the formula:
V = Ed
where:
V is the voltage between the plates,
E is the electric field strength, and
d is the distance between the plates.
In this case, we already know that the electric field strength (E) is 4500 V/m and the distance between the plates (d) is 10.0 cm.
First, let's convert the distance to meters:
10.0 cm = 10.0 cm * (1 m / 100 cm) = 0.10 m
Now, we can calculate the voltage between the plates:
V = Ed
V = (4500 V/m) * (0.10 m)
V = 450 V
Therefore, the voltage between the plates is 450 V.