Answer
64.8k+ views
Hint: In this, we first acquire the relation between the force, the charges, and the distance between the charges. Then further, we will give the vector form of the coulomb’s law. Understanding the basic concepts of coulomb’s law will help us find its limitations.
Formula used:
$F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{r^2}}}$
Complete step-by-step solution:
The Coulomb's law of electrostatic force states that the electrostatic force of attraction and repulsion are directly proportional to the product of the charges and inversely proportional to the square of the distance between charges.
$ \Rightarrow F\alpha {q_1} \times {q_2}$
$ \Rightarrow F\alpha \dfrac{1}{{{r^2}}}$
Where ${q_1}$ , ${q_2}$ are the charges, and $r$ is the distance between the charges.
Thus, the electrostatic force is given by,
$ \Rightarrow F\alpha \dfrac{{{q_1}{q_2}}}{{{r^2}}}$
$ \Rightarrow F = k\dfrac{{{q_1}{q_2}}}{{{r^2}}}$
Where k is constant of proportionality
The vector form of the coulomb’s law is known as:
$ \Rightarrow {\vec F_{12}} = k\dfrac{{{q_1}{q_2}}}{{{r_{12}}^2}}{\hat r_{12}}$
Here, ${r_{12}}$ is the displacement from charge $1$ to charge $2$ .
Force on the charge $2$ is given as:
$ \Rightarrow {\vec F_{21}} = k\dfrac{{{q_1}{q_2}}}{{{r_{21}}^2}}{\hat r_{21}}$
The magnitude of the force between charge $1$ and charge $2$ is equal.
$ \Rightarrow {\vec F_{12}} = {\vec F_{21}}$
Additional information:
1. There are some restrictions of this law as following:
2. This law is valid only for the point charges at rest.
3. Coulomb’s law can only be applied in the cases where the inverse square law is followed.
4. It is also tough to apply coulomb’s law where charges are in arbitrary shape because in such cases we cannot decide the distance between the two charges.
This law cannot be used directly to evaluate the charges of the big planets.
Note: The things to be at your fingertips for further information on solving these types of problems are: A tip to understand this type of concept is, to remember the scalar and vector quantities. When both the quantities, that is, the magnitude and the direction are involved, then those quantities will be vector quantities, but not all.
Formula used:
$F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{r^2}}}$
Complete step-by-step solution:
The Coulomb's law of electrostatic force states that the electrostatic force of attraction and repulsion are directly proportional to the product of the charges and inversely proportional to the square of the distance between charges.
$ \Rightarrow F\alpha {q_1} \times {q_2}$
$ \Rightarrow F\alpha \dfrac{1}{{{r^2}}}$
Where ${q_1}$ , ${q_2}$ are the charges, and $r$ is the distance between the charges.
Thus, the electrostatic force is given by,
$ \Rightarrow F\alpha \dfrac{{{q_1}{q_2}}}{{{r^2}}}$
$ \Rightarrow F = k\dfrac{{{q_1}{q_2}}}{{{r^2}}}$
Where k is constant of proportionality
The vector form of the coulomb’s law is known as:
$ \Rightarrow {\vec F_{12}} = k\dfrac{{{q_1}{q_2}}}{{{r_{12}}^2}}{\hat r_{12}}$
Here, ${r_{12}}$ is the displacement from charge $1$ to charge $2$ .
Force on the charge $2$ is given as:
$ \Rightarrow {\vec F_{21}} = k\dfrac{{{q_1}{q_2}}}{{{r_{21}}^2}}{\hat r_{21}}$
The magnitude of the force between charge $1$ and charge $2$ is equal.
$ \Rightarrow {\vec F_{12}} = {\vec F_{21}}$
Additional information:
1. There are some restrictions of this law as following:
2. This law is valid only for the point charges at rest.
3. Coulomb’s law can only be applied in the cases where the inverse square law is followed.
4. It is also tough to apply coulomb’s law where charges are in arbitrary shape because in such cases we cannot decide the distance between the two charges.
This law cannot be used directly to evaluate the charges of the big planets.
Note: The things to be at your fingertips for further information on solving these types of problems are: A tip to understand this type of concept is, to remember the scalar and vector quantities. When both the quantities, that is, the magnitude and the direction are involved, then those quantities will be vector quantities, but not all.
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