Gauss's Law

1 Mark Questions

1. Consider two hollow concentric spheres <math>S_1</math> and <math>S_2</math> enclosing charges 20 and 40

respectively, as shown in the figure, (i) Find out the ratio of the electric flux through them, (ii) How will the electric flux through the spheres S₁ change if a medium of dielectric constant e_r is introduced in the space inside S₁ in place of air? Deduce the necessary expression. [All India 2014] 4Q 2Q <math>S_2</math>

Ans.

1. According to Gauss' theorem,

<math display="block">\phi = \frac{\Sigma q}{} \propto \Sigma q</math> <math>\epsilon_0 \epsilon</math>

<math display="block">\frac{\phi_{S_1}}{\phi_{S_2}} = \frac{2Q}{2Q + 4Q} = \frac{1}{3}</math>

1. If the medium is filled in <math>S_1</math>, then

<math display="block">\phi_{S_1} = \frac{\Sigma q}{\varepsilon_0 \varepsilon_r} = \frac{2Q}{\varepsilon_0 \varepsilon_r}</math>

1. Two charges of magnitudes -20 and <math>+0</math> are located at points (a, 0) and (4a, 0),

respectively. What is the electric flux due to these charges through a sphere of radius 3a with its centre at the origin? [All India 2013]

Ans.

Gauss' theorem states that the total electric flux linked with closed surface S is

<math display="block">\phi_E = \phi \mathbf{E} \cdot d\mathbf{S} = \frac{q}{E_0}</math>

where, q is the total charge enclosed by the closed Gaussian (imaginary) surface.

-2Q <math>+Q</math> (0,0) (a,0) (3a,0) (4a,0)

The sphere enclose charge = -2Q

Therefore, <math>\phi = \frac{2Q}{\varepsilon_0}</math> (inwards)

(1) 3.A charge q is placed at the centre of a cube of side L. What is the electric flux passing through each face of the cube? [All India 2010; Foreign 2010] Ans.

By Gauss' theorem, total electric flux linked with a closed surface is given by

<math display="block">\phi = \frac{q}{\varepsilon_0}</math>

where, q is the total charge enclosed by the closed surface.

<math>\therefore</math> Total electric flux linked with cube, <math>\phi = \frac{q}{q}</math>

As charge is at centre, therefore, electric flux is symmetrically distributed through all 6 faces.

Flux linked with each face = <math>\frac{1}{6} \phi</math>

<math display="block">=\frac{1}{6}\times\frac{q}{\varepsilon_0}=\frac{q}{6\,\varepsilon_0}</math> (1)