Question

7.
Two metallic spheres of radii a and b respectively
are charged and joined by a conducting wire. Find
the ratio of electric field on the surface of the
spheres. (Separation between spheres is much
larger than their radii).
(1) alb
(2) bla
(3) a216²
(4) b-la?​

Answer 1

Answer:

the ratio of electric field on the surface of the  spheres is b : a.

Explanation:

If two spheres are connected by a wire then the electric potential of the two sphere must be same

So we will have

$V_1 = V_2$

$V_1 = \frac{kQ_1}{a}$

$V_2 = \frac{kQ_2}{b}$

now we know that electric field at the surface of the conductor is given as

$E = \frac{kQ}{R^2}$

now the ratio of electric field on the surface of two conductors is given as

$\frac{E_1}{E_2} = \frac{\frac{kQ_1}{a^2}}{\frac{kQ_2}{b^2}}$

Now we know that

$\frac{kQ_1}{a} = \frac{kQ_2}{b}$

so we have

$\frac{E_1}{E_2} = \frac{b}{a}$

#Learn

Topic : properties of conductors

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