Two plates of area 5.00 × 10-3 m2 are separated by a distance of 1.50 × 10-4 m. If a charge separation of 6.40 × 10-8 C is placed on the two plates, calculate the potential difference (voltage) between the two plates. Assume that the separation distance is small in comparison to the diameter of the plates.
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To calculate the potential difference between the two plates, we can use the formula:
V = Q / C
Where:
V = Potential difference (voltage)
Q = Charge separation on the plates
C = Capacitance
The capacitance is given by:
C = ε₀ * A / d
Where:
ε₀ = Permittivity of free space (8.85 × 10^-12 C^2/N·m^2)
A = Area of each plate
d = Distance between the plates
Let's calculate the capacitance first:
C = (8.85 × 10^-12 C^2/N·m^2) * (5.00 × 10^-3 m^2) / (1.50 × 10^-4 m)
C ≈ 2.96 × 10^-7 F
Now we can calculate the potential difference:
V = (6.40 × 10^-8 C) / (2.96 × 10^-7 F)
V ≈ 0.216 V
Therefore, the potential difference between the two plates is approximately 0.216 V.
To calculate the potential difference (voltage) between the two plates, you can use the formula for capacitance (C) and rearrange it to solve for voltage (V). The formula for capacitance is:
C = Q / V
Where:
C is the capacitance (in Farads, F)
Q is the charge (in Coulombs, C)
V is the voltage (potential difference, in Volts, V)
Rearranging the formula, we have:
V = Q / C
Given:
Area of each plate (A) = 5.00 × 10^(-3) m^2
Separation between plates (d) = 1.50 × 10^(-4) m
Charge separation (Q) = 6.40 × 10^(-8) C
To find the capacitance (C), we can use the formula:
C = ε₀ * (A / d)
Where:
ε₀ is the permittivity of free space, approximately 8.85 × 10^(-12) F/m
Plugging in the values, we have:
C = 8.85 × 10^(-12) F/m * (5.00 × 10^(-3) m^2 / 1.50 × 10^(-4) m)
C = 29.5 F
Now we can substitute the values of Q and C into the formula:
V = (6.40 × 10^(-8) C) / (29.5 F)
V ≈ 2.17 V
Therefore, the potential difference (voltage) between the two plates is approximately 2.17 volts.