Question

The ratio of the forces between two small spheres with constant charges (a) in air (b) in a medium of dielectric constant k is respectively

Answer 1

Let constant charges q appears each small sphere and distance between them is r.
you should know that force depends on medium, it decreases when we put two charges from air to a medium.
because permittivity of medium is directly Proportional to dielectric constant of medium.
e.g., $\epsilon_m=\epsilon_0K$ where, $\epsilon_m$ is permittivity of medium, K is dielectric constant of medium.

from Coulomb's law, in air medium force is given by, $F_{air}=\frac{1}{4\pi\epsilon_0}\frac{q^2}{r^2}$......(1)

In a dielectric medium of dielectric constant k is given.

force , $F_d=\frac{1}{4\pi\epsilon_0k}\frac{q^2}{r^2}$...... (2)

from equations (1) and (2),

$\frac{F_{air}}{F_d}=\frac{K}{1}$

hence, ratio of the forces is K : 1

Answer 2

Answer:K:1

Explanation:

Secondary SchoolPhysics 5+3 pts

The ratio of the forces between two small spheres with constant charges (a) in air (b) in a medium of dielectric constant k is respectively

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abhi178

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Let constant charges q appears each small sphere and distance between them is r.

you should know that force depends on medium, it decreases when we put two charges from air to a medium.

because permittivity of medium is directly Proportional to dielectric constant of medium.

e.g., \epsilon_m=\epsilon_0K where, \epsilon_m is permittivity of medium, K is dielectric constant of medium.

from Coulomb's law, in air medium force is given by, F_{air}=\frac{1}{4\pi\epsilon_0}\frac{q^2}{r^2}......(1)

In a dielectric medium of dielectric constant k is given.

force , F_d=\frac{1}{4\pi\epsilon_0k}\frac{q^2}{r^2}...... (2)

from equations (1) and (2),

\frac{F_{air}}{F_d}=\frac{K}{1}

hence, ratio of the forces is K : 1