Aparallel-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)

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

what experimental evidence led scientists to change from the previous model to this one?

a few of the positive particles aimed at a gold foil seemed to bounce back.

the colors of light emitted from heated atoms had very specific energies.

experiments with water vapor showed that elements combine in specific proportions.

cathode rays were bent in the same way whenever a magnet was brought near them.