Question

0-2.
derive exprrssion of electric force and electric pressure on the surface of charging conductor​

Answer 1

Answer:

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Explanation:

The charge provided to a conductor is uniformly distributed on its surface. a repulsive force acts by the charge on the rest part from the the ...

Answer 2

Explanation:

the vicinity of a charged conductor having charge per unit surface area(sigma) the electric field is given as

E=Sigma/(2epsilonzero)

Thus force per unit area=electric pressure

=qE

=(sigma )(sigma/2epsilonzero) N/m^2

=( sigma^2)/(2epsilonzero) N/m^2

Here (sigma) is surface charge density (C/m^2) and (epsilon zero) is permittivity of free space.

In the above expression for E , it has been considered in the case when charge of one surface of conductor is taken into account. For charge is taken on both the surfaces, the electric field is double that of the above value.

Thus electric pressure ie force per unit area is (takeE=2×sigma/2epsilonzero=sigma/epsilonzero)

P=(1/2)(epsilonzero)E^2