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<Ch20 Electric Fields and Forces-2 Item 7 Review | Constants Two identical closely spaced circular disks form a parallel-plate capacitor. Transferring 2.5x10 electrons from one disk to the other causes the electric field strength between them to be 1.6x105 N/C Part A What are the diameters of the disks? Express your answer with the appropriate units. HA Du Value Units Submit Request Answer Provide Feedback

Answer 1

Sol::

Given ::
Q= 2.5*10^9
E= 1.6*10^5 N/C

We know that capacitance C is

C = ε*A/d

The charge on a capacitor is:

Q = C*V
= ( ε*A/d )*( E*d )
Q = ε*A*E

Where

A = π*r^2
= π*D^2/4
As

A= Q/( ε*E )
π*D^2/4= Q/( ε*E )

where "D" represents the plate diameter.

D^2 = 4*Q/( π*ε*E )

D = sqrt( 4*Q/( π*ε*E )
  
=sqrt[4*(2.5*10^9)*(1.6*10^-19) / ( π* (8.854*10^-12) * 1.6*10^5 ) ]

D = 18.96*10^-3 m
≈ 19 mm

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