A pair of parallel conducting plates each 40cm x 50cm have a separation of 5mm and carry charges of ±5CµC. Calculate the potential difference between the plates.
I have a problem on how to get the permittivity of the dielectric too.
I have a problem on how to get the permittivity of the dielectric too.
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There is only enough information to derive the potential difference OR the permittivity but not both independently.
So we must assume the dielectric is air, with permittivity εo.
Plate area A = 0.4m*0.5m = 0.2m^2
So capacitance is C = εo*A / d = 8.85*10^-12 * 0.2 / 0.005 F = 3.54*10^-10 F
Potential difference is therefore
V = Q / C = 0.5*10^-6 C / 3.54*10^-10 F
V = 1410 Volts
So we must assume the dielectric is air, with permittivity εo.
Plate area A = 0.4m*0.5m = 0.2m^2
So capacitance is C = εo*A / d = 8.85*10^-12 * 0.2 / 0.005 F = 3.54*10^-10 F
Potential difference is therefore
V = Q / C = 0.5*10^-6 C / 3.54*10^-10 F
V = 1410 Volts