Question

Two point charges ${{q}_{2}}=3\times {{10}^{-6}}C$ and ${{q}_{1}}=5\times {{10}^{-6}}C$ are located at (3, 5, 1) and (1, 3, 2) m respectively. Find $\overrightarrow{{{F}_{12}}}$ and $\overrightarrow{{{F}_{21}}}$ using vector form of coulomb’s law. Also, find their magnitudes.

Answer 1

First write down the position vectors of the two charges with the help of the given data and then calculate the vector $\overrightarrow{{{r}_{12}}}$ to calculate the value of $\overrightarrow{{{F}_{12}}}$. We also need to calculate the unit vector along $\overrightarrow{{{r}_{12}}}$. Then we can perform similar steps for calculating $\overrightarrow{{{F}_{21}}}$.

Formula used:
$\overrightarrow{{{F}_{12}}}=\dfrac{k{{q}_{1}}{{q}_{2}}}{{{r}_{12}}}\widehat{{{r}_{12}}}$
where $\overrightarrow{{{F}_{12}}}$ is the electrostatic force exerted on charge 1 by charge 2, k is a proportionality constant, ${{q}_{1}}$, ${{q}_{2}}$ are the charges, ${{r}_{12}}$ is the distance between the charges and $\widehat{{{r}_{12}}}$ is unit vector along the vector $\overrightarrow{{{r}_{12}}}$.
$\overrightarrow{{{r}_{12}}}=\overrightarrow{{{r}_{1}}}-\overrightarrow{{{r}_{2}}}$
where $\overrightarrow{{{r}_{1}}}$ and $\overrightarrow{{{r}_{2}}}$ are position vectors of the charges.

Complete step by step answer:
The position vector of the first charge (${{q}_{1}}$) is $\overrightarrow{{{r}_{1}}}=\left( 1\widehat{i}+3\widehat{j}+2\widehat{k} \right)m$.
And the position vector of the second charge (${{q}_{2}}$) is $\overrightarrow{{{r}_{2}}}=\left( 3\widehat{i}+5\widehat{j}+1\widehat{k} \right)m$.
Then, $\overrightarrow{{{r}_{12}}}=\overrightarrow{{{r}_{1}}}-\overrightarrow{{{r}_{2}}}$
$\Rightarrow \overrightarrow{{{r}_{12}}}=\left( 1\widehat{i}+3\widehat{j}+2\widehat{k} \right)-\left( 3\widehat{i}+5\widehat{j}+1\widehat{k} \right)=(1-3)\widehat{i}+(3-5)\widehat{j}+(2-1)\widehat{k}$
$\Rightarrow \overrightarrow{{{r}_{12}}}=-2\widehat{i}-2\widehat{j}+\widehat{k}$
The magnitude of the vector $\overrightarrow{{{r}_{12}}}$ is equal to ${{r}_{12}}=\sqrt{{{(-2)}^{2}}+{{(-2)}^{2}}+{{(1)}^{2}}}=3m$.
Therefore, the unit vector along $\overrightarrow{{{r}_{12}}}$ is $\widehat{{{r}_{12}}}=\dfrac{\overrightarrow{{{r}_{12}}}}{{{r}_{12}}}=\dfrac{-2\widehat{i}-2\widehat{j}+\widehat{k}}{3}$
Therefore, the electrostatic force or Coulomb’s force on the charge 1 exerted charge 2 is equal to $\overrightarrow{{{F}_{12}}}=\dfrac{k{{q}_{1}}{{q}_{2}}}{{{r}_{12}}}\widehat{{{r}_{12}}}$ …. (i)
The value of constant k in vacuum is given to be $k=9\times {{10}^{9}}N{{m}^{2}}{{C}^{-2}}$
Now, substitute the known values in equation (i).
$\Rightarrow \overrightarrow{{{F}_{12}}}=\dfrac{9\times {{10}^{9}}\times 3\times {{10}^{-6}}\times 5\times {{10}^{-6}}}{3}\times \dfrac{-2\widehat{i}+-2\widehat{j}+\widehat{k}}{3}$
$\Rightarrow \overrightarrow{{{F}_{12}}}=45\times {{10}^{-3}}\left( \dfrac{-2\widehat{i}+-2\widehat{j}+\widehat{k}}{3} \right)N$
We know that the magnitude of a unit vector is unity. Therefore, the magnitude of the electrostatic force on charge 1 exerted by charge 2 is equal to ${{F}_{12}}=45\times {{10}^{-3}}N$.
Similarly, we will find that $\overrightarrow{{{r}_{21}}}=2\widehat{i}+2\widehat{j}-\widehat{k}$ and ${{r}_{21}}=3m$.
Therefore, $\widehat{{{r}_{21}}}=\dfrac{\overrightarrow{{{r}_{21}}}}{{{r}_{21}}}=\dfrac{2\widehat{i}+2\widehat{j}-\widehat{k}}{3}$.
Therefore, the electrostatic force or Coulomb’s force on the charge 2 exerted charge 1 is equal to $\overrightarrow{{{F}_{21}}}=\dfrac{k{{q}_{1}}{{q}_{2}}}{{{r}_{21}}}\widehat{{{r}_{21}}}$ …. (ii)
Substitute the known values in equation (ii).
$\Rightarrow \overrightarrow{{{F}_{12}}}=\dfrac{9\times {{10}^{9}}\times 3\times {{10}^{-6}}\times 5\times {{10}^{-6}}}{3}\times \dfrac{2\widehat{i}+2\widehat{j}-\widehat{k}}{3}$
$\therefore \overrightarrow{{{F}_{12}}}=45\times {{10}^{-3}}\left( \dfrac{2\widehat{i}+2\widehat{j}-\widehat{k}}{3} \right)N$

With this, we get that the magnitude of the electrostatic force on charge 2 exerted by charge 1 is equal to ${{F}_{21}}=45\times {{10}^{-3}}N$.

Note: Electrostatic force between two charges is action reaction force like gravitational force. This means that the force exerted by charge 1 is equal in magnitude with the force exerted by charge 2 on charge 1 but the direction of the two forces opposite. We can consider the given question as an example of this.