Physics Ncert Solutions Class 12th Physics Class 12th Electrostatic Potential and Capacitance

2.15 A spherical conducting shell of inner radius r1 and outer radius r2 has a charge Q.

(a) A charge q is placed at the centre of the shell. What is the surface charge density on the inner and outer surfaces of the shell?

(b) Is the electric field inside a cavity (with no charge) zero, even if the shell is not spherical, but has any irregular shape? Explain.

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alok kumar singh | Contributor-Level 10

3 months ago

2.15 Charge placed at the centre of the shell is +q. Hence a charge of magnitude of –q will be induced to the inner surface of the shell. Therefore, the total charge on the inner surface of the shell is –q.

Surface charge density at the inner surface of the shell is given by the relation,

$σ_{1} = \frac{T o t a l c h a r g e}{I n n e r s u r f a c e a r e a} = \frac{- q}{4 π r_{1}^{2}}$ …………(1)

When a charge of magnitude of Q is placed on the outer shell and the radius of outer shell is $r_{2}$ . The total charge experienced by the outer shell will be Q + q. The surface charge density at the outer surface of the shell is given by

$σ_{2} = \frac{T o t a l c h a r g e}{O u t e r s u r f a c e a r e a} = \frac{Q + q}{4 π r_{2}^{2}}$ …………(2)

The electric field

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Surface charge density at the inner surface of the shell is given by the relation,

$σ_{1} = \frac{T o t a l c h a r g e}{I n n e r s u r f a c e a r e a} = \frac{- q}{4 π r_{1}^{2}}$ …………(1)

$σ_{2} = \frac{T o t a l c h a r g e}{O u t e r s u r f a c e a r e a} = \frac{Q + q}{4 π r_{2}^{2}}$ …………(2)

The electric field intensity inside a cavity is zero, even if the shell is not spherical and has any irregular shape.

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2.15 Charge placed at the centre of the shell is +q. Hence a charge of magnitude of –q will be induced to the inner surface of the shell. Therefore, the total charge on the inner surface of the shell is –q.Surface charge density at the inner surface of the shell is given by the relation,<math><msub><mrow><mrow><mi>σ</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub><mo>=</mo><mi></mi><mfrac><mrow><mrow><mi>T</mi><mi>o</mi><mi>t</mi><mi>a</mi><mi>l</mi><mi></mi><mi>c</mi><mi>h</mi><mi>a</mi><mi>r</mi><mi>g</mi><mi>e</mi></mrow></mrow><mrow><mrow><mi>I</mi><mi>n</mi><mi>n</mi><mi>e</mi><mi>r</mi><mi></mi><mi>s</mi><mi>u</mi><mi>r</mi><mi>f</mi><mi>a</mi><mi>c</mi><mi>e</mi><mi></mi><mi>a</mi><mi>r</mi><mi>e</mi><mi>a</mi></mrow></mrow></mfrac><mo>=</mo><mfrac><mrow><mrow><mo>-</mo><mi>q</mi></mrow></mrow><mrow><mrow><mn>4</mn><mi>π</mi><msubsup><mrow><mrow><mi>r</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msubsup></mrow></mrow></mfrac></math> …………(1)When a charge of magnitude of Q is placed on the outer shell and the radius of outer shell is <math><msub><mrow><mrow><mi>r</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub></math> . The total charge experienced by the outer shell will be Q + q. The surface charge density at the outer surface of the shell is given by<math><msub><mrow><mrow><mi>σ</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub><mo>=</mo><mi></mi><mfrac><mrow><mrow><mi>T</mi><mi>o</mi><mi>t</mi><mi>a</mi><mi>l</mi><mi></mi><mi>c</mi><mi>h</mi><mi>a</mi><mi>r</mi><mi>g</mi><mi>e</mi></mrow></mrow><mrow><mrow><mi>O</mi><mi>u</mi><mi>t</mi><mi>e</mi><mi>r</mi><mi></mi><mi>s</mi><mi>u</mi><mi>r</mi><mi>f</mi><mi>a</mi><mi>c</mi><mi>e</mi><mi></mi><mi>a</mi><mi>r</mi><mi>e</mi><mi>a</mi></mrow></mrow></mfrac><mo>=</mo><mfrac><mrow><mrow><mi>Q</mi><mo>+</mo><mi>q</mi></mrow></mrow><mrow><mrow><mn>4</mn><mi>π</mi><msubsup><mrow><mrow><mi>r</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msubsup></mrow></mrow></mfrac></math> …………(2)The electric field intensity inside a cavity is zero, even if the shell is not spherical and has any irregular shape.

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