Coulomb's Law

Electricity and Magnetism Level 4

What will be the force on charge $q_2$ due to another charge $q_1$ separated by distance $r$ in a medium with relative permittivity $\epsilon_r$ ?

Details and Assumptions

$\epsilon_0$ is permittivity of vacuum
$\hat{r_{12}}$ is unit vector in the direction of $q_1$ with respect to $q_2$ and vice-versa.

\frac{1}{4\pi\epsilon_0}×\frac{q_1q_2}{r^2}\hat{r_{12}}

\frac{1}{4\pi\epsilon_r}×\frac{q_1q_2}{r^2}\hat{r_{12}}

\frac{1}{4\pi\epsilon_0\epsilon_r}×\frac{q_1q_2}{r^2}\hat{r_{21}}

\frac{1}{4\pi\epsilon_r}×\frac{q_1q_2}{r^2}\hat{r_{21}}

\frac{1}{4\pi\epsilon_0}×\frac{q_1q_2}{r^2}\hat{r_{21}}

\frac{1}{4\pi\epsilon_0\epsilon_r}×\frac{q_1q_2}{r^2}\hat{r_{12}}

2 solutions

Pranjal Jain
Mar 3, 2015

The answer seems to be $\frac{1}{4\pi\epsilon_0\epsilon_r}×\frac{q_1q_2}{r^2}\hat{r_{21}}$ but note that this is net force on $q_2$ ie it includes force by $q_1$ and force by medium. However, force asked in problem is force due to $q_1$ only which is $\frac{1}{4\pi\epsilon_0}×\frac{q_1q_2}{r^2}\hat{r_{21}}$ (independent of medium)

To tell you the truth, I did not know that. Thanks a lot.

Raghav Vaidyanathan - 6 years, 3 months ago

That's nice! Keep B'ant-ing and resharing! $\ddot\smile$

Pranjal Jain - 6 years, 3 months ago

Even I didn"t know it.

Atul Solanki - 6 years ago

Sorry, I couldn't get what you are trying to say. The force is electrostatic right? So, the relative permittivity has to be taken into account.

The presence of a dielectric medium does affect the electrostatic Coulumb's force between two charges (kept in that medium).

help me clear this doubt

thanks

Soumo Mukherjee - 6 years, 2 months ago

If you were asked to calculate the Gravitational force on earth due to The Sun if the space between the Earth and The Sun is uniformly (or non uniformly) filled with small masses, then what would be your answer?

Purushottam Abhisheikh - 6 years, 1 month ago

Ok. I have got it. I forgot that I left this unmentioned here.

Btw... Thanks for replying

Soumo Mukherjee - 6 years ago

I feel the dumbest amongst here as i did the vector part wrong though i knew the fact that it is independent of the medium 😭😭😭

Jitender Sharma - 2 years, 1 month ago

Purushottam Abhisheikh
Mar 3, 2015

The electric force on $q_2$ due to $q_1$ (pay attention on the meaning and don't take it as net force on $q_2$ ) is simply independent of medium present around it. So the answer is what it comes in a perfect vacuum medium i.e. $\frac{1}{4\pi\epsilon_0}×\frac{q_1q_2}{r^2}\hat{r_{21}}$ .

Yours is more precise than Pranjal's

Muhammad Arifur Rahman - 6 years, 3 months ago

Coulomb's Law

2 solutions

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