Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Try this! Two identical conducting spheres A and B, carry equal charge. They are separatedby a distance much larger than their diameters, and the force between them isF. A third identical conducting sphere, C is uncharged. Sphere C is first touched t

Two identical conducting spheres A and B, carry equal charge. They are separated
by a distance much larger than their diameters, and the force between them is
F. A third identical conducting sphere, C is uncharged. Sphere C is first touched to
A, then to B, and then removed. As a result, the force between A and B would
be equal to :

  • Option 1)

    F

  • Option 2)

    3F/4

  • Option 3)

    3F/8

  • Option 4)

    F/2

 

Answers (3)

best_answer

As we have learned

 

Coulombic force -

F\propto Q_{1}Q_{2}=F\propto \frac{Q_{1}Q_{2}}{r^{2}}=F=\frac{KQ_{1}Q_{2}}{r^{2}}

- wherein

K - proportionality Constant 

Q1 and Q2 are two Point charge

 

 

 

Initially force = F 

 let charge on both sphere = Q 

when in contact , charge will transfer till potential become equal , hence Q/2 on each sphere

When in contact with another sphere then change will transfer till charge on both is 34/4

 

new force    \alpha Q_{1},Q_{2}

 =3F/8


Option 1)

F

This is incorrect

Option 2)

3F/4

This is incorrect

Option 3)

3F/8

This is correct

Option 4)

F/2

This is incorrect

Posted by

Avinash

View full answer

JEE Main high-scoring chapters and topics

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

(D) is the correct option as 

Lets us consider the first case when A comes in contact with C Charge transfer occurs till an equilibrium is attained as shown in the following steps.Let A have a charge Q as well as B have a charge Q and Qa,Qb,Qc be the new charges acquired after C comes in contact with A and B and r be radius of A,Band c

Va=Vc

(1/4 pi e0) Qa/r  = (1/4 pi e0) Qc/r

By solving the above equation we would the the following result

Qc=Q*r/(r+r)

Qc=Q/2

Similarly in the second case  Qc=Q/2 in the end both A and B have charges of Q/2 each by Coloumb's inverse square law

F=1/(4 pi e0) Q^2/r^2------->(1)

New force F"=1/(4 pi e0)Q^2/4r^2----------->(2)

Dividing (1) and (2) F/F"=1/4

Therfore new force is F/4 as seen in D option

Posted by

Samanyu

View full answer

let Q be the initial charge on the spheres. 

  •  Force  F=kQ^2/r^2   

 when sphere C is attached to sphere A then net charge on the both spheres(A & C) is   =(Q+0)/2 =Q/2 

Now when sphere C containing Q/2 charge is attached with sphere B (containing Q charge)

            the net charge on the spheres B & C =(Q/2+Q)/2 =3Q/4   

then force between A & B   F1=(kQ/2.3Q/4)/r^2

                                           =(kQ^2/r^2).(3/8)

                                           =3F/8

                                          that is the answer

Posted by

mohammad zeeshan

View full answer

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025 results out; paper 2 BArch, BPlanning scorecard next, session 2 registration open
anam-khan

Only 1 female among JEE Main 2025 session 1 toppers; 12 out of 14 from general category
anam-khan

Forgot JEE Main password? Here’s how to reset for checking results on jeemain.nta.nic.in

Latest Question