The magnitude of the electrostatic force between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them.
If two point charges $q_{1}$, $q_{2}$ are separated by a distance r in vacuum, the magnitude of the force (F) between them is given by
If $\epsilon _{r}$ be the permittivity of any medium then $K=\frac{\epsilon _{r}}{\epsilon _{0}}$ where K is dielectric constant of the medium.
Coulomb force is a vector quantity. If position vector of $q_{1}$ is $\vec{r_{1}}$ and $q_{2}$ is $\vec{r_{2}}$ then relative position vector of $q_{1}$ with respect to $q_{2}$ is $\vec{r_{12}}=\vec{r_{1}} -\vec{r_{2}}$ then the coulomb force on $q_{1}$ by $q_{2}$ will be
If two point charges $q_{1}$, $q_{2}$ are separated by a distance r in vacuum, the magnitude of the force (F) between them is given by
\[\large F=K\frac{q_{1}q_{2}}{r^{2}}\]
where K is the proportionality constant. In SI System $\large K=\frac{1}{4\pi \epsilon _{0}}$ and $\epsilon _{0}$ is the permittivity in free space $\epsilon _{0}=8.854 \times 10^{-12} \ C^{2}\ N^{-1}\ m^{-2}$If $\epsilon _{r}$ be the permittivity of any medium then $K=\frac{\epsilon _{r}}{\epsilon _{0}}$ where K is dielectric constant of the medium.
Coulomb force is a vector quantity. If position vector of $q_{1}$ is $\vec{r_{1}}$ and $q_{2}$ is $\vec{r_{2}}$ then relative position vector of $q_{1}$ with respect to $q_{2}$ is $\vec{r_{12}}=\vec{r_{1}} -\vec{r_{2}}$ then the coulomb force on $q_{1}$ by $q_{2}$ will be
\[\vec{F_{12}}=\frac{1}{4\pi \epsilon _{0}}\frac{q_{1}q_{2}}{\left |\vec{r_{12}} \right |^{2}}\hat{r_{12}}\]
And relative position vector of $q_{2}$ with respect to $q_{1}$ is $\vec{r_{21}}=\vec{r_{2}} -\vec{r_{1}}$ then the coulomb force on $q_{2}$ by $q_{1}$ will be
\[\vec{F_{21}}=\frac{1}{4\pi \epsilon _{0}}\frac{q_{1}q_{2}}{\left |\vec{r_{21}} \right |^{2}}\hat{r_{21}}\]
For n number of charges
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| Coulomb Force |
Remember $[\vec{F_{12}}$ and $[\vec{F_{21}}$ are equal in magnitude but opposite in direction
Coulomb's Force due to Multiple Charges:
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| Coulomb's Force due to Multiple Charges |
Forces on $q_{1}$ charge are $\vec{F_{12}}$, $\vec{F_{13}}$ and $\vec{F_{14}}$ therefore the resultant charge $\vec{F}= \vec{F_{12}}+\vec{F_{13}}+\vec{F_{14}}$
\[\therefore \vec{F}= \frac{1}{4\pi \epsilon _{0}}\frac{q_{1}q_{2}}{\left | r_{12} \right |^{2}}\hat{r_{12}}+\frac{1}{4\pi \epsilon _{0}}\frac{q_{1}q_{3}}{\left | r_{13} \right |^{2}}\hat{r_{13}}+\frac{1}{4\pi \epsilon _{0}}\frac{q_{1}q_{4}}{\left | r_{14} \right |^{2}}\hat{r_{14}}\]
\[\vec{F}= \frac{1}{4\pi \epsilon _{0}}\sum_{j=2}^{n}\frac{q_{1}q_{j}}{\left | r_{1j} \right |^{2}}\hat{r_{1j}}\]
