Question

Two circular plates with radius 6.22 cm have equal and opposite charge of magnitude 0.952 Calculate the electric field between the two plates. Assume that the separation distance is small in comparison to the diameter of the plates. Number N/ C The plates are slowly pulled apart, doubling the separation distance. What changes occur with the electric field between the plates? (Again, assume the separation distance remains small in comparison to the diameter of the plates.) O The electric field increases by a factor of 4. O The electric field increases by a factor of 2. O The electric field stays the same. O The electric field decreases by a factor of 2 O The electric field decreases by a factor of 4.

Answer 1

Acooording to the concept of the electric potential and capacitance

the electric field b/w the parallel plate E=Q/2e A

Given that

charge q=0.952*10^-6 C

radius r=0.0622 m

Area =3.14*0.0622^2=1.215*10^-2 m^2

now we find the electric field

electric field E=0.952*10^-6/2*8.85*10^-12*1.215*10-2

=4.43*10^6 N/c

now we find the factor of the electric field

Eelectric field in parallel plates E=q/eA

here we the electric fielld is related with only area of the plate not the distance b/w the plates

so if the distance b/w the plates the electric field does not change