Question

electric field match the following​

Answer 1

Answer:

ok

Explanation:

ANSWER

a ) Electric field outside sphere, take a spherical gaussion surface

E×4πr

2

=

ξ

0

V

E=

4πξ

0

r

2

V

∴E ∝

r

2

1

b ) Now E=

r

2

KV

we know V=−∫

d

r

Edr=−KV[

r

1

]

r

=

r

KV

c ) Apply Gauss law taking a sphere at a distance r

we get E×4πr

2

=

3

4

πr

3

ξ

0

p

E=

2ξ

Pr

∴E∝r

d ) Shell Theorem : electric potential inside a changed conducting sphere is constant and is equal to the value at the surface

∴V=constant