A solid sphereical conductor of radius R has a spherical cavity of radius a (a < R) at its centre. A charge + Q is kept at the centre. The charge at the inner surface, outer and at a position r (a < r < R) are respectively(a) + Q, – Q, 0(b) – Q, + Q, 0(c) 0, – Q, 0(d) + Q , 0, 0
Answers
Answer:
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.
Answer:
at inner surface - 0 , at outer surface -0 and at at a position r - + Q
Explanation:
due to charge distribution ,the will be even distributed on surface of spherical conductor of radius R and so there will be no charge on inside and outside of the shell
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