Question

In a cylindrical capacitor with plate radii a and b, the potential difference between plates is v.The electric field between the plates at a radial distance r from axis is​

Answer 1

Answer:

Explanation:

Radii of the plate in cylindrical capacitor = a and b (Given)

Potential difference between plates = v (Given)

Let the inner sphere has a charge +Q  and outer sphere has -Q

Thus, between the sphere we have

Potential of the plate of radius a  -

Q=CV

V= (Q/C)

Or E = (KQ/r²)

= KQ (b-a)/ (ab)

Q = (Vab)/ (K (b-a))

E = (K/r²) [Vab/ (K (b-a))

= (abV)/ (r² (b-a))

Answer 2

Answer:

V/R ln(b/a)

Explanation:

The above upload answer is for spherical capacitor and here it is asked for Cylindrical Capacitor.

Electric Field Between Plates is Lamba/2 pi epsilon r.

Lambda=Q×A(LINEAR CHARGE DENSITY)

Substituting Lambda,

Q=CV

Capacitance=2 pi epsilon/ln (b/a)

Substituting Q value

Hence Answer is V/R ln(b/a)

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Question from Irodov,HC Verma.

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