Two large horizontal metal plates are separated by 4mm. The lower plate is at a potential of (-)6v.What potential should be applied to the upper plate to create field strength of 4000v/m upwards in the space between the plates?(a)(+)22(b)(+)10(c)(-)10(d)(-)22
Can any one tell me the solution of this mcq
To find the potential that should be applied to the upper plate to create a field strength of 4000 V/m upwards in the space between the plates, we can use the formula for electric field:
Electric field (E) = Voltage (V) / Distance between the plates (d)
Given:
Distance between the plates (d) = 4 mm = 0.004 m
Electric field (E) = 4000 V/m
Rearranging the formula:
Voltage (V) = Electric field (E) * Distance between the plates (d)
Now, substituting the given values:
Voltage (V) = 4000 V/m * 0.004 m
Voltage (V) = 16 V
Since the lower plate is at a potential of -6 V, the potential to be applied to the upper plate would be the sum of the potential difference (16 V) and the potential of the lower plate (-6 V):
Potential of the upper plate = -6 V + 16 V = 10 V
Therefore, the potential that should be applied to the upper plate is (+)10 V.
The correct answer is option (b): (+)10.
To find the potential that should be applied to the upper plate, we need to use the formula for electric field strength (E) between parallel plates:
E = V/d,
where E is the electric field strength, V is the potential difference between the plates, and d is the distance between the plates.
In this case, we have the following known values:
Electric field strength (E) = 4000 V/m,
Distance between the plates (d) = 4 mm = 0.004 m,
Potential of the lower plate (V) = -6 V.
Using the formula, we can rearrange it to solve for the potential difference (V):
V = E * d.
Plugging in the values, we have:
V = (4000 V/m) * (0.004 m) = 16 V.
Since the potential of the lower plate is -6 V, the potential of the upper plate will be:
Potential of the upper plate = Potential of the lower plate + Potential difference = -6 V + 16 V = 10 V.
Therefore, the potential that should be applied to the upper plate to create a field strength of 4000 V/m upwards is (+)10 V. The correct answer is (b) (+)10.