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A parallel-plate air capacitor has a capacitance of 920 pF. The charge on each plate is 3.90 μC. What is the potential difference between the plates? Express your answer with the appropriate units. Part B If the charge is kept constant, what will be the potential difference between the plates if the separation is doubled? Express your answer with the appropriate units. Part C How much work is required to double the separation? Express your answer with the appropriate units (mJ)

Respuesta :

Answer: a)235.9*10^-6 V; b) 471.8 *10^-6 V; c) 1.08*10^-13 J

Explanation: In order to explain this problem we have to use the expression for a capacitor od two parallel plates, which is given by:

C=Q/V where Q and V are the charge and voltage difference in teh capacitor, respectively.

Then we have;

V=Q/C=920*10^-12/(3.9*10^-6)= 235.9*10^-6 V

If we double the separation between the plates, the potencial difference will be:  ( Q is constant)

Vnew=Q/Cnew  where Cnew is given by;

Cnew=ε *A/(2*d) = 1/2*(ε *A/d)=1/2* Co ; then:

Vnew=2*Q/(Co)=2*Vo=2*4.24 *10^3 V=471.8 *10^-6 V

Finally, the work made oin the capacitor is the difference of the final and initial potential energy so:

Uo=Q^2/(2Co)=

Uf= Q^2/(2Cnew)               Cnew=Co/2

W=Uf-Uo= Q^2(1/Co)-Q^2(1/2*Co)=Q^2(1/2*Co)=1.08*10^-13 J

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