Question: Q. 8. A point charge +Q is placed at the centre O of an uncharged hollow spherical conductor of inner radius ’ a ’ and outer radius ’ b ‘. Find the following:

(i) The magnitude and sign of the charge induced on the inner and outer surface of the conducting shell.

(ii) The magnitude of electric field vector at a distance (i) r=a2, and (ii) r=2b, from the centre of the shell.

A&E [CBSE SQP 2017-18]

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Solution:

Ans. (i) As the electrostatic field inside a conductor is zero, using Gauss’s law, charge on the inner surface of the shell =Q

Charge on the outer surface of the shell =+Q1/2

(ii) To show using Gauss’s law

Expression for electric field for radius, r=a2 :

E=14πε04Qa2

Expression for electric field for radius, r=2b :

(1)E=14πε0Qb2

[CBSE Marking Scheme, 2017]

Detailed Answer :

(i) Charge placed at the centre of a shell is +Q. Hence, a charge of magnitude Q will be induced to the inner surface of the shell. Therefore, total charge on the inner surface of the shell is Q.

A charge of +Q is induced on the outer surface of the shell, so a charge of magnitude Q is placed on the outer surface of the shell.

(ii) The electric flux is electric field times the area of spherical surface

ϕ=EA=4Eπr2=Qε0

So electric field outside the sphere (r>R) is seen to be identical to that of a point charge Q at center of sphere.

E=Q4πε0r2

Magnitude of electric field vector at a distance

r=a2 E=14πε04Qa2

Magnitude of electric field vector at a distance

r=2b E=14πε04Q4b2=14πε0Qb21



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