In two cases, two identical conducting spheres are given equal charges, in one case of the same type whereas in another case of opposite type. The distance between the spheres is not large comparing with the diameter. Let F 1 and F 2  be the magnitude of the force of interaction between the spheres, as shown, then

 

Option 1 -

F 1 > F 2

Option 2 -

F 1 = F 2

Option 3 -

F 1 < F 2

Option 4 -

information is not sufficient to draw the conclusion

0 1 View | Posted 2 weeks ago
Asked by Shiksha User

  • 1 Answer

  • R

    Answered by

    Raj Pandey | Contributor-Level 9

    2 weeks ago
    Correct Option - 3


    Detailed Solution:

    In Case I when both are positively charged, due to induction positive charge moves outwards on spheres, increasing effective distance between centres of charge causing magnitude of the force to decrease.

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S
Syed Aquib Ur Rahman

The medium is affects force. We know this from the constant, k, in Coulomb's Law, that depends on the medium. If you place a charge in an insulator or dielectric medium, like water, the force between them will decrease. This decrease can be taken into account by the medium's permittivity. 

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Syed Aquib Ur Rahman

The polarity of the charges that are interacting help us understand the direction. Like charges, whether it's both positive or negative, will always repel. Unlike charges, when one is positive and the other is negative, will always attract each other. The force acts along the line connecting the two charges. 

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Syed Aquib Ur Rahman

The main condition would be that the point charges have to be at rest or be in a stationary position, for Coulomb's Law to work. The thing is, if the charge particles move, we have to consider the impact of magnetic forces then. 

A
alok kumar singh

According to question, we can write

F C A = F C B = K q Q x 2 + ( d 2 ) 2 F = 2 F C A c o s θ = 2 K q Q x [ x 2 + ( d 2 ) 2 ] 3 2                         

For maxima of force  d F d x = 0 , s o  

x = d 2 2

 

R
Raj Pandey

mg In equilibrium, Fe=T sin θ

mg=T cos θ

 tan θ=Femg=q24π0x2×mg

also  tan θs?=x/2l

Hence, x 2 l = q 2 4π0x2×mg

x 3 = 2 q 2 p 4πε0mg

x=(q2l2π0mg)1/3

Therefore xl1/3

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