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What is the force acting between two point charges?
Answer
401.7k+ views
Hint: There was an experiment conducted by Coulomb, in 1785. He gave a statement to explain what the force acting between two point charges is. A point charge is a charge assumed to be at a geometric point. It has no dimensions. Here we will start by assuming 2 charges separated by a distance r and then derive COULOMB'S LAW.
Complete answer:
We will learn about a very interesting topic now called COULOMB’S LAW.
This law states that the force of attraction or repulsion between 2 point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of distance between them, and acts along the line joining the 2 charges.
Let’s derive this,
Let there be 2 point charges a and b separated by distance r, look at the following image:
Hence applying the law we get: \[F \propto ab\] and \[F \propto \dfrac{1}{{{r^2}}}\]
\[F \propto \dfrac{{ab}}{{{r^2}}}\]
Or, \[F = k\dfrac{{ab}}{{{r^2}}}\]
Where k is constant of proportionality.
\[k = \dfrac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9}N{m^2}{C^{ - 2}}\] and
\[{\varepsilon _0} = 8.854 \times {10^{ - 12}}{N^{ - 1}}{m^{ - 2}}{C^2}\].
Thus force between two point charges is written as:
\[F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{ab}}{{{r^2}}}\].
It can also be represented in vector form as:
If force is on b due to a,
\[{\vec F_{21}} = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{ab}}{{r_{21}^2}}{\hat r_{21}}\]
And 2) if force is on a due to b,
\[{\vec F_{12}} = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{ab}}{{r_{12}^2}}{\hat r_{12}}\]
Where \[{\hat r_{12}}\,and\,{\hat r_{21}}\] are the unit vectors from a to b and b to a respectively.
Note: The unit vectors are mentioned to tell that the force can be from object a to b or from object b to a. It is always better to mention the vector quantities because it specifies direction along with magnitude. Also remember that Coulomb’s law is valid in a vast range \[{10^{ - 17}}m\] to \[{10^7}m\]and it is not affected by introducing any other charges.
Complete answer:
We will learn about a very interesting topic now called COULOMB’S LAW.
This law states that the force of attraction or repulsion between 2 point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of distance between them, and acts along the line joining the 2 charges.
Let’s derive this,
Let there be 2 point charges a and b separated by distance r, look at the following image:
![seo images](https://www.vedantu.com/question-sets/8a9a9274-dca5-4251-a138-db53e4ea50d42302482733712256823.png)
Hence applying the law we get: \[F \propto ab\] and \[F \propto \dfrac{1}{{{r^2}}}\]
\[F \propto \dfrac{{ab}}{{{r^2}}}\]
Or, \[F = k\dfrac{{ab}}{{{r^2}}}\]
Where k is constant of proportionality.
\[k = \dfrac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9}N{m^2}{C^{ - 2}}\] and
\[{\varepsilon _0} = 8.854 \times {10^{ - 12}}{N^{ - 1}}{m^{ - 2}}{C^2}\].
Thus force between two point charges is written as:
\[F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{ab}}{{{r^2}}}\].
It can also be represented in vector form as:
If force is on b due to a,
\[{\vec F_{21}} = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{ab}}{{r_{21}^2}}{\hat r_{21}}\]
And 2) if force is on a due to b,
\[{\vec F_{12}} = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{ab}}{{r_{12}^2}}{\hat r_{12}}\]
Where \[{\hat r_{12}}\,and\,{\hat r_{21}}\] are the unit vectors from a to b and b to a respectively.
Note: The unit vectors are mentioned to tell that the force can be from object a to b or from object b to a. It is always better to mention the vector quantities because it specifies direction along with magnitude. Also remember that Coulomb’s law is valid in a vast range \[{10^{ - 17}}m\] to \[{10^7}m\]and it is not affected by introducing any other charges.
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