Question

A small sphere of radius a carrying a
positive charge q is placed concentrically
inside a large hollow conducting shell of
radius b (b>a). This outer shell has charge
Q on it. Show that if these spheres are
connected by a conducting wire, charge will
always flow from the inner sphere to the
outer sphere irrespective of the magnitude
of the two charges. All India 2009​

Answer 1

Answer:

Explanation:

Potential of the Shell:

V2=k(q1+q2)r2V2=k(q1+q2)r2

Potential of the Sphere:

V1=kq1r1+kq2r2V1=kq1r1+kq2r2

As r1<r2,V1>V2r1<r2,V1>V2. So positive charge will flow from sphere(inside) to the shell(outside) when connected.

Answer 2

Answer:

The charge on the inner sphere is greater towards the outer sphere.

V1 > V2 ,while b > a

Explanation:

• There is a Sphere of radius a,surrounded by a sphere of radius b.The centre is common for both spheres.
• The potential of inner sphere is V1 and outer sphere is V2.
• The potential difference V = V1 - V2

$v1 = \frac{kq1}{b} + \frac{kq2}{b}$

$v2 = \frac{kq2}{a} + \frac{kq1}{b}$

$v = \frac{kq1}{b} + \frac{kq2}{b} - \frac{kq2}{a} - \frac{kq1}{b}$

$v = kq2( \frac{1}{b} - \frac{1}{a})$

$v = kq2( \frac{a - b}{ab} )$

If b>a,then

$v = (- ve)$

Thus,the charge on inside sphere is greater than the outside sphere.

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