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Two balls with equal charges are in a vessel with ice at -10 C at a distance of 25 cm from each other. On forming water at 0 C, the balls are brought nearer to 5cm for interaction between them to be same. If dielectric constant of water at 0 C is 80, the dielectric constant of ice at -10 C is ?
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Two balls with equal charges are in a vessel with ice at -10 C at a di...
Calculation of Dielectric Constant of Ice at -10 C


Given Data:


  • Initial distance between two balls = 25 cm

  • Final distance between two balls = 5 cm

  • Dielectric constant of water at 0 C = 80



Formula:

For two charged spheres, the force of interaction between them is given by:

F = (Q^2 / 4πεr^2)

where:


  • F = force of interaction between the spheres

  • Q = charge on each sphere

  • ε = dielectric constant of the medium between the spheres

  • r = distance between the centers of the spheres



Solution:

Let the initial charge on each sphere be Q1 and the final charge on each sphere be Q2. Since the balls have equal charges, Q1 = Q2.

Using the formula for force of interaction between the spheres, we have:

F1 = (Q1^2 / 4πε1r1^2)

F2 = (Q2^2 / 4πε2r2^2)

Since the force of interaction between the spheres is the same in both cases, we can equate F1 and F2:

(Q1^2 / 4πε1r1^2) = (Q2^2 / 4πε2r2^2)

Since Q1 = Q2, we can simplify the equation to:

(1 / ε1r1^2) = (1 / ε2r2^2)

Substituting the given values, we have:

(1 / (εice)(25)^2) = (1 / (80)(5)^2)

Solving for εice, we get:

εice = (80)(5)^2(25)^2 = 25000


Answer:

The dielectric constant of ice at -10 C is 25000.
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Two balls with equal charges are in a vessel with ice at -10 C at a distance of 25 cm from each other. On forming water at 0 C, the balls are brought nearer to 5cm for interaction between them to be same. If dielectric constant of water at 0 C is 80, the dielectric constant of ice at -10 C is ?
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