Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Two uniformly charged spherical conductors  and  of radii <img al

Two uniformly charged spherical conductors \mathrm{A} and \mathrm{B} of radii \mathrm{5 \mathrm{~mm}} and \mathrm{10 \mathrm{~mm}} are separated by a distance of \mathrm{ 2\mathrm{~cm}.} If the spheres are connected by a conducting wire, then in equilibrium condition, the ratio of the magnitudes of the electric fields at the surface of the sphere \mathrm{ A} and \mathrm{ B} will be :

Option: 1

1: 2


Option: 2

2: 1


Option: 3

1: 1


Option: 4

1: 4


Answers (1)

best_answer

When the spheres are connated by conducting wire, their potential will be same

\mathrm{V_1=V_2 }

\mathrm{\frac{K Q_1}{R_1}=\frac{K Q_2}{R_2} }

\mathrm{\frac{Q_1}{Q_2}=\frac{R_1}{R_2}=\frac{1}{2}}

Electic field on the surface of the sphere

\mathrm{ E=\frac{K Q}{R^2} }

\mathrm{ \frac{E_1}{E_2}=\frac{Q_1 R_2^2}{Q_2 R_1^2}=\frac{2}{1} }

Electric field on the surface of the one sphere is independent of other. (as separation is large as compare to size of sphere)

Hence (2) is correct option.

Posted by

Ajit Kumar Dubey

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

In JEE Main 2025 April 3 shift 2, chemistry takes U-turn to be tougher than physics, mathematics
anam-khan

JEE Main 2025 April 3 Exam Analysis 2025: Shift 1 paper easy to moderate; chemistry section ‘easiest’
anam-khan

JEE Mains 2025 admit card out for April 7, 8, 9 exams; download on jeemain.nta.nic.in

Latest Question