Question

A solid spherical conductor of radius R has a spherical cavity
of radius aſa<R) at its centre. A charge + is kept at the
center. The charge at the inner surface, outer and at a
position r(a<r <R) are respectively ​

Answer 1

Consider a Gaussian sphere of radius a just including the inner surface of the sphere.

Since the surface is inside the conducting region,

E

=

0

Thus, flux through the gaussian surface is zero.

Using Gauss' Law, ∮E.ds=

ϵ

o

q

enclosed

We get the charge enclosed within the Gaussian sphere is zero as E=0.

But, the cavity contains a charge of +Q.

Hence, the inner surface of the conductor contains a charge of −Q.

Now, consider a Gaussian sphere of radius r (a<r<R) as shown in the figure.

The surface of the sphere lies within the conducting spherical shell.

Hence

E

=

0

Thus, flux through the Gaussian surface is zero.

By Gauss' law, the total charge enclosed in the Gaussian surface is zero.

But, the the charge inside the cavity is +Q and the inner surface of the conductor has a charge of −Q.

Hence, charge at the location r is zero.

Initially, the charge on the conductor is zero.

Hence by principle of conservation of charge, total charge on outer surface and inner surface and the interior is zero.

Inner surface has a charge of −Q and the interior has zero charge.

Hence, the outer surface has a charge of +Q.

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Answer 2

Answer:

The inner surface of cavity will be −Q due to induction. At outer surface, the charge will be +Q and at a point between P at a position r(a<r<R) will be zero.

Explanation:

Consider a Gaussian sphere of radius a just including the inner surface of the sphere.

Since the surface is inside the conducting region,

→     →

E=   0

Thus, flux through the gaussian surface is zero.

Using Gauss' Law,   ∮E.ds= $\frac{ q_{enclosed} }{ϵo}$

​We get the charge enclosed within the Gaussian sphere is zero as E=0.

But, the cavity contains a charge of +Q.

Hence, the inner surface of the conductor contains a charge of −Q.

Now, consider a Gaussian sphere of radius r  (a<r<R) as shown in the figure.

The surface of the sphere lies within the conducting spherical shell.

Hence  

E=0

Thus, flux through the Gaussian surface is zero.

By Gauss' law, the total charge enclosed in the Gaussian surface is zero.

But, the the charge inside the cavity is +Q and the inner surface of the conductor has a charge of −Q.

Hence, charge at the location r is zero.

Initially, the charge on the conductor is zero.

Hence by principle of conservation of charge, total charge on outer surface and inner surface and the interior is zero.

Inner surface has a charge of −Q and the interior has zero charge.

Hence, the outer surface has a charge of +Q.

Reference Link

• https://brainly.in/question/11264291
• https://brainly.in/question/47918121