Question

Two identical metal spheres having equal and similar charges repel each other with a force of $103N$when they are placed $10cm$apart in a medium of dielectric constant $5$. Determine the charge on each sphere.

Answer 1

The electrostatic force is directly proportional to the magnitude of both charges and inversely proportional to the distance between them. This proportionality is removed by introducing a term known as the permittivity of free space. However, in different mediums it has different values, the ratio of permittivity of the substance to that of free space is called the dielectric constant.

Complete step by step answer::
The electrostatic force between two charged bodies when the dielectric constant of a material is given, is written as-
$F = \dfrac{1}{{4\pi {\varepsilon _0}K'}}\dfrac{{{q_1}{q_2}}}{{{r^2}}}$
Where F is the force between charges.
${q_1}$and ${q_2}$ are the charges which exert the force on each other.
r is the distance between these charges.
$\dfrac{1}{{4\pi {\varepsilon _0}}}$is the Coulomb’s constant, it is sometimes also represented as $K$. The term ${\varepsilon _0}$is known as the permittivity of free space.
$K'$ is the dielectric constant of a material.
In the question, it is said that the metal spheres have equal and similar charges,
Therefore,
${q_1} = {q_2} = q$
The distance between the metal spheres (charges) is, $r = 10cm$
In meters, $r = 0.1m$ \left\\{ {1m = 100cm} \right\\}
The value of the coulomb’s constant is, $K = 9 \times {10^9}$.
The dielectric constant of the medium is, $K' = 5$
The electrostatic force between the metal spheres is, $F = 103N$
Putting all these values in the formula of force, we have-
$F = \dfrac{{K{q^2}}}{{K'{r^2}}}$
$\Rightarrow 103 = \dfrac{{9 \times {{10}^9} \times {q^2}}}{{5 \times {{\left( {0.1} \right)}^2}}}$
$\Rightarrow {q^2} = \dfrac{{515 \times {{10}^{ - 2}} \times {{10}^{ - 9}}}}{9}$
$\Rightarrow {q^2} = 5.72 \times {10^{ - 10}}$
$\Rightarrow q = 2.39 \times {10^{ - 5}}$
$\Rightarrow q = 23.9 \times {10^{ - 6}}C$or $23.9\mu C$

Note: The dielectric constant is defined as the ratio of permittivity in a medium- ${\varepsilon _r}$to the permittivity in free space- ${\varepsilon _0}$, it can be represented by- $K' = \dfrac{{{\varepsilon _r}}}{{{\varepsilon _0}}}$ or ${\varepsilon _r} = K'{\varepsilon _0}$. To calculate the electrostatic force when the charges are placed inside a medium, the permittivity of the medium is used. If here, we know the value of the dielectric constant$K'$, we can easily write the formula as$F = \dfrac{1}{{4\pi {\varepsilon _0}K'}}\dfrac{{{q_1}{q_2}}}{{{r^2}}}$.