Calculate maximum values of voltage, charge and force for the following capacitor.
A parallel plate capacitor with capacitance C = 45*10^6 F has a distance d = 1 mm between the plates. The space between the plates is filled with paper with a disruptive field strength of Edis = 1.5*10^7 V/m .
What potential difference Vmax can be applied across the plates without creating a spark? What is the maximal charge Qmax which can be stored by the capacitor without creating a spark and if the potential difference Vmax is applied to the capacitor, what is the force between the plates?
1 Answer | Add Yours
We have a parallel plate capacitor with capacitance C = 45*10^-6 F and the plates are 1 mm apart. There is paper between the plates. The paper has a Edis of 1.5*10^7 V/m. The maximum voltage field that the paper can withstand is 1.5*10^7 V/m. As the distance between the plates is just 1 mm, the maximum voltage that can be applied between them is 1/1000 of 1.5*10^7
=> Vmax = 1.50*10^4 V
We know that Q = C*V
As C = 45*10^-6 F
Qmax = C*Vmax
=> Qmax = 45*10^-6*1.50*10^4 C
=> Qmax = 67.50*10^-2 C
The electric field can be expressed in terms of N/C, Qmax= 67.50*10^-2 C
Fmax = 1.5*10^7*67.5*10^-2
=> 10.12*10^5 N
=> 1.01 MN
The required values are Vmax = 1.50*10^4 V, Qmax = 67.50*10^-2 C and Fmax = 1.01 MN
We’ve answered 317,511 questions. We can answer yours, too.Ask a question