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
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Hint For solving this question we will use Coulomb’s law which is nothing but the relation between and charge on bodies and corresponding force generated. Far along we will see the force on two identical spheres divided by some distance. Keeping that in mind we will solve this problem.
Formula used- The force between two charges is specified by
$F = \dfrac{{k{q_1}{q_2}}}{{{d^2}}}$
where ${q_1}$ and ${q_2}$ are the two charges respectively and $d$is the distance between them.
Complete step-by-step solution
Coulomb’s law states that the electrical force between two charged objects is directly proportional to the product of the quantity of charge on the objects and inversely proportional to the square of the parting distance between the two objects.
Its scalar form, the law is:
$F = \dfrac{{k{q_1}{q_2}}}{{{d^2}}}$
where $k$ is Coulomb’s constant $\left( {k \simeq 9 \times {{10}^9}N{m^2}{C^{ - 2}}} \right)$,
${q_1}$ and ${q_2}$ are the signed magnitude of the charges, and
the scalar $d$ is the distance between the charges.
Since the charge on both spheres is similar and equal say $q$ and kept in a medium having a dielectric constant say $k'$.
Therefore the force between them is given by the above formula,
$F = \dfrac{{k{q^2}}}{{k'{d^2}}}$ ................. $\left( 1 \right)$
Now the values given in the question are as follows:
The force between the two identical charged metal spheres is,
$F = 103N$
Distance between the charged spheres is,
$d = 10cm = 0.1m$
The dielectric constant of the medium in which the two charged spheres are kept is,
$k' = 5$
Now substitute all the values in the equation $\left( 1 \right)$ we get,
$F = \dfrac{{k{q^2}}}{{k'{d^2}}}$
$ \Rightarrow 103 = \dfrac{{9 \times {{10}^9} \times {q^2}}}{{5 \times {{\left( {0.1} \right)}^2}}}$
Taking the ${q^2}$ on the left-hand side of the equation we get,
${q^2} = \dfrac{{103 \times 5 \times 0.01}}{{9 \times {{10}^9}}}$
$ \Rightarrow {q^2} = 572.22 \times {10^{ - 12}}$
Taking square root both the sides we get,
$q = 23.92 \times {10^{ - 6}}C$
Hence, the charge on each metallic sphere is $23.92 \times {10^{ - 6}}C$ .
Note The SI derived unit of electric charge is the coulomb, which is defined as an ampere second whereas the force has the unit, Newton. Correspondingly, two positive charges repel each other and the positive and negative charges attract each other or in common like charges repel each other and unlike charges attract each other.
Formula used- The force between two charges is specified by
$F = \dfrac{{k{q_1}{q_2}}}{{{d^2}}}$
where ${q_1}$ and ${q_2}$ are the two charges respectively and $d$is the distance between them.
Complete step-by-step solution
Coulomb’s law states that the electrical force between two charged objects is directly proportional to the product of the quantity of charge on the objects and inversely proportional to the square of the parting distance between the two objects.
Its scalar form, the law is:
$F = \dfrac{{k{q_1}{q_2}}}{{{d^2}}}$
where $k$ is Coulomb’s constant $\left( {k \simeq 9 \times {{10}^9}N{m^2}{C^{ - 2}}} \right)$,
${q_1}$ and ${q_2}$ are the signed magnitude of the charges, and
the scalar $d$ is the distance between the charges.
Since the charge on both spheres is similar and equal say $q$ and kept in a medium having a dielectric constant say $k'$.
Therefore the force between them is given by the above formula,
$F = \dfrac{{k{q^2}}}{{k'{d^2}}}$ ................. $\left( 1 \right)$
Now the values given in the question are as follows:
The force between the two identical charged metal spheres is,
$F = 103N$
Distance between the charged spheres is,
$d = 10cm = 0.1m$
The dielectric constant of the medium in which the two charged spheres are kept is,
$k' = 5$
Now substitute all the values in the equation $\left( 1 \right)$ we get,
$F = \dfrac{{k{q^2}}}{{k'{d^2}}}$
$ \Rightarrow 103 = \dfrac{{9 \times {{10}^9} \times {q^2}}}{{5 \times {{\left( {0.1} \right)}^2}}}$
Taking the ${q^2}$ on the left-hand side of the equation we get,
${q^2} = \dfrac{{103 \times 5 \times 0.01}}{{9 \times {{10}^9}}}$
$ \Rightarrow {q^2} = 572.22 \times {10^{ - 12}}$
Taking square root both the sides we get,
$q = 23.92 \times {10^{ - 6}}C$
Hence, the charge on each metallic sphere is $23.92 \times {10^{ - 6}}C$ .
Note The SI derived unit of electric charge is the coulomb, which is defined as an ampere second whereas the force has the unit, Newton. Correspondingly, two positive charges repel each other and the positive and negative charges attract each other or in common like charges repel each other and unlike charges attract each other.
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