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  • (a) Using Gauss' law, Obtain expressions for the electric field (i) inside, and (ii) outside a positively charged spherical shell. (b) Show graphically variation of the electric field as a function of the distance r from the centre of the

(a) Using Gauss' law, Obtain expressions for the electric field (i) inside, and (ii) outside a positively charged spherical shell.

(b) Show graphically variation of the electric field as a function of the distance r from the centre of the sphere.

(c) A square plane sheet of side 10cm is inclined at an angle of 30^{\circ} with the direction of a uniform electric field of 200 NC-1. Calculate the electric flux passing through the sheet.

 

 

 

 

 
 
 
 
 

Answers (1)

(i) inside the shell

Since there is no charge enclosed in inside the shell.

By Gauss's law flox \phi = E\times 4\pi r^{2}

                           E\times 4\pi r^{2}=0

Therefore E = 0

Thus electric field inside the shell is zero

(ii) Out of the shell

We know,

\sigma = uniform surface charge density of spherical shell

Total flux = E\times 4\pi r^{2}

r= radius

The spherical shell is an equipotential surface so the electric field on each point is the same.

In spherical shell the electric field E and area element as are parallel.

Here the total charge, q =\sigma \times 4\pi r^{2}

By Gauss's law

E\times 4\pi r^{2}=\frac{\sigma }{\epsilon _{0}}\times 4\pi r^{2}

E= \frac{\sigma R^{2}}{\varepsilon _{0}r^{2}}= \frac{q}{4\pi R^{2}}\times \frac{R^{2}}{\varepsilon _{0}r^{2}}

Therefore, E= \frac{1}{4\pi \varepsilon _{0}}\frac{q}{r^{2}}

This is same as the electric field produced by a point charge.

(b) We got two conditions in the above case

 That is

The electric field inside the spherical shell is zero

The electric field outside the spherical shell is 

E= \frac{1}{4\pi \varepsilon _{0}}\frac{q}{r^{2}}

which means E\propto \frac{1}{r^{2}}

By this, we can show it graphically

From this,

when r < R electric field is zero (inside the shell)

r > R electric field is inversely proportional to the resistance

(c) Given 

 r = 10 cm = 0.1m

Electric field 'E' = 200 NC-1

The angle between the sheet and electric field - 30^{\circ}

The angle between the electric field and normal to the plane sheet \theta =90 -30=60^{\circ}

Electric flux \phi= E.\Delta S

                        = E.\Delta S\cos \theta

                        = E(0.1)^{2}\cos 60^{\circ}

                        = 1\frac{Nm^2}{C}

Posted by

Safeer PP

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