
Two identical metal spheres possess $+60\;C$ and $-20\;C$ of charges. They are bought in contact and then separated by $10\;cm$. The force between them is
\[\begin{align}
& A.36\times {{10}^{13}}N \\
& B.36\times {{10}^{14}}N \\
& C.36\times {{10}^{12}}N \\
& D.3.6\times {{10}^{12}}N \\
\end{align}\]
Answer
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Hint: We know that the electric force is force experienced due to a pair charges being given by Coulomb's force. Clearly, the electric force depends on the charge q and inversely proportional to the square of the distance between them. Since the charges are brought to contact, there is some rearrangement of the charges. We can use the above to solve the given problem.
Formula used: $E=\dfrac{q_{1}q_{2}}{4\pi\epsilon_{0}r^{2}}$
Complete step by step answer:
We know that electric force is the force between the given charges. This is given by Coulomb's law. From the coulomb's law, we know that, the electrical force $F$ between two charges $q_{1}$ and $q_{2}$ which is separated at a distance $r$ is given as
$F\propto q_{1}q_{2}$ and $F\propto\dfrac{1}{r^{2}}$
Combining the two we have,$\implies F\propto \dfrac{q_{1}q_{2}}{r^{2}}$
We can remove the proportionality, using a constant, then we have the coulomb's law of electric force is given as $F=k\dfrac{q_{1}q_{2}}{r^{2}}$,where $k=\dfrac{1}{4\pi\epsilon_{0}}$ which is a constant with a value $9\times 10^{9}$ .
Given that two charges $q_{1}=+60C$ and $q_{2}=-20$ which is brought to contact, then the net charge is given as $Q=60-20=40C$. This net charge gets distributed equally.
This charge now gets separated at a distance $r=10cm$, then, we can say that $Q_{1}=20C$ and $Q_{2}=20C$
Then the force between them is given as $F= \dfrac{9\times 10^{9}\times 20\times 20}{({10^{-1}})^2}=36\times 10^{13}N$
So, the correct answer is “Option A”.
Note: We know that the electric force due to a pair of charges is given by Coulomb's law. This force can be attractive or repelling depending on the nature of the charges. An electric field can be produced by a time-varying electric field or an electrical charge. These can be either attracting or repelling in nature.
Formula used: $E=\dfrac{q_{1}q_{2}}{4\pi\epsilon_{0}r^{2}}$
Complete step by step answer:
We know that electric force is the force between the given charges. This is given by Coulomb's law. From the coulomb's law, we know that, the electrical force $F$ between two charges $q_{1}$ and $q_{2}$ which is separated at a distance $r$ is given as
$F\propto q_{1}q_{2}$ and $F\propto\dfrac{1}{r^{2}}$
Combining the two we have,$\implies F\propto \dfrac{q_{1}q_{2}}{r^{2}}$
We can remove the proportionality, using a constant, then we have the coulomb's law of electric force is given as $F=k\dfrac{q_{1}q_{2}}{r^{2}}$,where $k=\dfrac{1}{4\pi\epsilon_{0}}$ which is a constant with a value $9\times 10^{9}$ .
Given that two charges $q_{1}=+60C$ and $q_{2}=-20$ which is brought to contact, then the net charge is given as $Q=60-20=40C$. This net charge gets distributed equally.
This charge now gets separated at a distance $r=10cm$, then, we can say that $Q_{1}=20C$ and $Q_{2}=20C$
Then the force between them is given as $F= \dfrac{9\times 10^{9}\times 20\times 20}{({10^{-1}})^2}=36\times 10^{13}N$
So, the correct answer is “Option A”.
Note: We know that the electric force due to a pair of charges is given by Coulomb's law. This force can be attractive or repelling depending on the nature of the charges. An electric field can be produced by a time-varying electric field or an electrical charge. These can be either attracting or repelling in nature.
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