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Academic session 2026-27, Chapter 2 in the current curriculum.
Welcome to the NCERT Solutions for Class 12 Physics - Chapter Electrostatic Potential and Capacitance. This page offers a step-by-step solution to the specific question from Exercise 1, Question 4:
A spherical conductor of radius 12 cm has a charge of 1.6 x 10-7C distributed uniformly on its surface. What is the electric field
(a) Inside the sphere
(b) Just outside the sphere
(c) At a point 18 cm from the centre of the sphere?
. With detailed answers and explanations for each chapter, students can strengthen their understanding and prepare confidently for exams. Ideal for CBSE and other board students, this resource will simplify your study experience.A spherical conductor of radius 12 cm has a charge of 1.6 x 10-7C distributed uniformly on its surface. What is the electric field
(a) Inside the sphere
(b) Just outside the sphere
(c) At a point 18 cm from the centre of the sphere?
(a) Radius of the spherical conductor, r = 12 cm = 0.12 m
Charge is uniformly distributed over the conductor, q = 1.6 × 10 -7 C
Electric field inside a spherical conductor is zero. This is because if there is field inside the conductor, then charges will move to neutralize it.
(b) Electric field E just outside the conductor is given by the relation,

Where,
= Permittivity of free space

Therefore, the electric field just outside the sphere is 105 N C-1
(c) Electric field at a point 18 m from the centre of the sphere = E1
Distance of the point from the centre, d = 18 cm = 0.18 m

Therefore, the electric field at a point 18 cm from the centre of the sphere is .
4.4 x 104 N/C
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Welcome to the NCERT Solutions for Class 12 Physics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 4: A spherical conductor of radius 12 cm has a charge of 1.6 x 10-7C distributed uniformly on its....
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Comments
This was really helpful.. Manish mannu has commented that the unit was given wrong here.. But actually it's correct.. Pls do check properly before commenting..
It can not be solved by using gauss law????
Thanku
Tnk u so much for ur help
It was very very helpful. Ty Saral studyðŸËŠðŸËÅ
nice