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the ratio of the forces between two small spheres charged to constant potentials in air and in a medium of dielectric constant is?

the ratio of the forces between two small spheres charged to constant potentials in air and in a medium of dielectric constant is?
 

Grade:12th pass

3 Answers

Eshan
askIITians Faculty 2095 Points
5 years ago
Dear student,

Forces acting between two charged spheres=\dfrac{1}{4\pi\epsilon}\dfrac{q_1q_2}{r^2}
Since the spheres are chargecd to same potential, the charges over them is also same.
Hence the force acting between them\propto \dfrac{1}{\epsilon}

Hence the ratio=\dfrac{F_{air}}{F_{medium}}=\dfrac{\epsilon_{medium}}{\epsilon_{air}}=\dfrac{\epsilon_r\epsilon_0}{\epsilon_0}=\epsilon_r
Arun
25750 Points
5 years ago
Dear Ishan

Force due to charge in air = > q1q2/(4*pi*e)r^2 = F1
Force due to charge in di – electric medium = F2
or
F2 = q1q2/(4*pi**e*K)r^2
where K is the dielectric constant.
Dividing the forces we get,
F1/ F2 = K
In case of any query, please feel free to ask.
 
Regards
Arun (askIITiana forum expert)
Aryan
13 Points
2 years ago
It has been given that potential on spheres is kept constant, let V
Also let charge on each spheres in air be Q
So the expression for the charge will be
 
V=Q/(4pi€R)
=> Q = 4pi€VR
Now since this potential is kept constant and medium is introduced, the charges on the spheres will change keeping potential constant ,let the new charges on each sphere be Q' , thus the expression for Q' will be
V=Q'/4pi€kR
=> Q'= 4pi€kRV
Hence 
F1=(Q)^2/(4pi€d^2)=(4pi€V^2 R^2)/d^2
Similarly
F2=(Q')^2/(4pi€kd^2)=(4pi€kV^2 R^2)/d^2
Thus 
F1/F2= 1/k

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