Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9

Q.

Two metal spheres, one of radius R and the other of radius 2R respectively have the same surface charge density σ. They are brought in contact and separated. What will be the new surface charge densities on them?

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya

a

σ1=56σ, σ2=52σ

b

σ1=52σ, σ2=53σ

c

σ1=53σ, σ2=56σ

d

σ1=52σ, σ2=56σ

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Detailed Solution

detailed_solution_thumbnail

when spheres are kept in contact, charge redistribution takes place to get same potential. 

Total combined charge = 20πR2σ

Potential of both the spheres become equal, hence Q = QI2 where Q is the charge on sphere of radius R and QI is the charge on sphere of radius 2R.

Thus charge on sphere of radius R becomes 20πR2σ3 , while charge on sphere of radius 2R becomes 40πR2σ3

Hence new charge densities will be

σ1=53σ, σ2=56σ

Watch 3-min video & get full concept clarity
score_test_img

courses

No courses found

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!

best study material, now at your finger tips!

  • promsvg

    live classes

  • promsvg

    progress tracking

  • promsvg

    24x7 mentored guidance

  • promsvg

    study plan analysis

download the app

gplay
mentor

Download the App

gplay
whats app icon
personalised 1:1 online tutoring