Two large horizontal, parallel metal plates are 2.0 cm apart in air and the upper is maintained at a positive potential relative to the lower so that field strength between them is 〖2.5×10〗^5 〖Vm〗^(-1).
Jhh6
To solve this problem, we can use the formula for electric field strength (E):
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
Given:
Potential difference (V) = ?
Distance between the plates (d) = 2.0 cm = 0.02 m
Electric field strength (E) = 2.5 x 10^5 V/m
We need to find the potential difference (V) between the plates. Rearranging the formula, we get:
V = E * d
Substituting the values:
V = (2.5 x 10^5 V/m) * 0.02 m
V = 5000 V
Therefore, the potential difference between the plates is 5000 V.
To find the potential difference between the two plates, you need to use the formula:
V = Ed
Where:
V is the potential difference (in volts)
E is the electric field strength (in volts per meter)
d is the distance between the plates (in meters)
In this case, you are given that the field strength (E) is 2.5 × 10^5 V/m and the distance (d) between the plates is 2.0 cm, which is equal to 0.02 m.
Now, substitute the given values into the formula:
V = (2.5 × 10^5 V/m) × (0.02 m)
Calculating this expression:
V = 5 × 10^3 V
Therefore, the potential difference between the two plates is 5,000 volts.