Question

3. A hollow spherical conductor of radius 12 cm is given
a charge of 2.4 x 10-° C. (i) Calculate the potential
on the surface of the conductor. (ii) Calculate the
intensity of electric field in air at a distance of 4 m
from the centre of the sphere.
Ans. (i) 180 V, (ii) 1.35 V-m .​

Answer 1

Answer :-

$(1)\implies \boxed{ \sf{V = 180 \: volt \:} } \\ (2 )\implies \boxed{ \sf{E =1.35\:\:V-m}} \:$

To Find :-

(1) Potential on the surface of the conductor.

(2) Electric field intensity at a distance of 4m from it's centre .

Explanation :-

Given that ;

$\implies \sf{ charge(q) = 2.4 \times {10}^{ - 9} c }\\ \implies \sf{radius \: (r) = 12 \: cm \: = 0.12 \: m} \\ \implies \sf{ a \: particular \: distance \: (d) = 4m}$

(1) Potential on surface (V)

We know that ,

$\implies \boxed{ \sf{V = \: \frac{kq}{r}} } \: \: \tt{on \: surface}\\ \\ \sf{ here \: k = 9 \times {10}^{9} \: } \because \: k = \frac{1}{4\pi\epsilon_{o}} \: \\ \\ \implies \sf{V = \frac{9 \times {10}^{9} \times 2.4 \times {10}^{ - 9} }{0.12} }\\ \\ \implies \sf{V = \frac{9 \times 24}{12} \times 10} \\ \\ \implies \boxed{ \sf{V = 180 \: volt \:} }$

(2) Electric field intensity at 4 m

we know that ,

$\to \boxed{ \sf{E = \frac{1}{4\pi\epsilon_{o}} \: \frac{q}{ {d}^{2} } }} \\ \therefore \\ \to \sf{E = \frac{9 \times {10}^{9} \times 2.4 \times {10}^{ - 9} }{16} } \\ \\ \to \sf{E = \frac{9 \times 24}{160} } \\ \\ \to \boxed{ \sf{E = 1.35\:\:V-m}}$

Extra :-

If we need electric field intensity inside the conductor so it will zero , because free charges inside the conductor is zero .

Answer 2

$\tt{\huge{\orange {Hello!!! }}} S.N.$

QUESTION :

A hollow spherical conductor of radius 12 cm is given a charge of 2.4 x 10-° C.

TO CALCULATE :

(i) Calculate the potential on the surface of the conductor.

(ii) Calculate the intensity of electric field in air at a distance of 4 m

from the centre of the sphere.

ANSWER :

Ans :

(i) 180 V,

(ii) 1.35 V-m .

SOLUTION :

We have the following information given :

Radius of Hollow Sphere = 0.12 m.

Charge = 24 × 10 ^ { - 9 } Coulumb.

V = { kQ } / { R }

Substituting the above values :

V = [ { 9 × 10 ^ 9 } × { 2.4 × 10 ^ { - 9 } ] / [ 0.12 ] = 18/ Volt.......(A1)

___________

E = [ 1 ] / [ 4 π e 0 ] × [ q ] / [ { d } ^ 2 ]

Substituting the Above Values :

E = [ [ 1 ] / [ 9 × 10 ^ { -9 } ] ] ×[ [ 2.4 × 10 ^ { - 9 } ] / [ { 0.12 } ^ 2 ] ]

=> E = 1.35 V × M.............(A2)