Electrostatic Potential Due to Charge Distribution at Specific Points

In this article, we will explore the concept of electrostatic potential due to a specific charge distribution in three-dimensional space. We will delve into the mathematical derivation to find the electrostatic potential at different points in relation to two charged particles located at specific positions.

Introduction to Electrostatic Potential

The electrostatic potential at a point in space is a measure of the potential energy per unit charge at that point. Mathematically, it is defined by the formula:

[ V(vec{r}) frac{1}{4piepsilon_0} int_{text{all space}} frac{rho(vec{r'})}{|vec{r} - vec{r'}|} d^3r' ]

Charge Distribution and its Influence on Electrostatic Potential

Consider two charges, ( q ) and ( -q ), located at the points ( (0, 0, -a) ) and ( (0, 0, a) ) respectively in a three-dimensional Cartesian coordinate system. Here, ( q ) is a positive charge and ( -q ) is a negative charge of equal magnitude.

Electrostatic Potential at the Point (0, 0, z)

To find the electrostatic potential at the point ( (0, 0, z) ), we use the formula for the potential due to a point charge:

[ V(vec{r}) frac{q}{4piepsilon_0 r} ]

Here, the distance ( r ) from the charge ( q ) at ( (0, 0, -a) ) to the point ( (0, 0, z) ) is:

[ r_1 sqrt{(0-0)^2 (0-0)^2 (z-(-a))^2} |z a| ]

Similarly, the distance ( r ) from the charge ( -q ) at ( (0, 0, a) ) to the point ( (0, 0, z) ) is:

[ r_2 sqrt{(0-0)^2 (0-0)^2 (z-a)^2} |z - a| ]

Calculating the Electrostatic Potential

The total electrostatic potential at ( (0, 0, z) ) due to the two charges is the sum of the individual potentials:

[ V(vec{r}) V_1(vec{r}) V_2(vec{r}) frac{q}{4piepsilon_0 (z a)} - frac{q}{4piepsilon_0 (z-a)} ]

By substituting ( epsilon_0 8.854 times 10^{-12} ) F/m (permittivity of free space), we get:

[ V(vec{r}) frac{q}{4pi (8.854 times 10^{-12}) (z a)} - frac{q}{4pi (8.854 times 10^{-12}) (z - a)} ]

After simplifying the expression, we obtain:

[ V(vec{r}) frac{q}{4pi times (8.854 times 10^{-12})} left( frac{1}{z a} - frac{1}{z - a} right) ]

Converting the constants, we get:

[ V(vec{r}) frac{q}{36pi times 10^{-9}} left( frac{1}{z a} - frac{1}{z - a} right) ]

To further simplify, we combine the terms inside the parenthesis:

[ V(vec{r}) frac{q}{36pi times 10^{-9}} left( frac{(z - a) - (z a)}{(z a)(z - a)} right) frac{q}{36pi times 10^{-9}} left( frac{-2a}{z^2 - a^2} right) ]

Simplifying the expression further, we obtain:

[ V(vec{r}) frac{-2qa}{18pi times 10^{-9}} frac{1}{z^2 - a^2} -frac{18qa}{10^9} frac{1}{z^2 - a^2} ]

This is the final expression for the electrostatic potential at ( (0, 0, z) ).

Electrostatic Potential at the Point (0, 0, 0)

To find the electrostatic potential at the point ( (0, 0, 0) ), we substitute ( z 0 ) into the potential formula:

[ V(0) -frac{18qa}{10^9} frac{1}{0^2 - a^2} -frac{18qa}{10^9} frac{1}{-a^2} frac{18qa}{10^9 a^2} frac{18q}{10^9 a} ]

However, the electrostatic potential at the midpoint ( (0, 0, 0) ) is actually zero because the contributions from the two charges ( q ) and ( -q ) cancel each other out:

[ V(0) frac{q}{4piepsilon_0 a} - frac{q}{4piepsilon_0 a} 0 ]

Conclusion

In summary, the electrostatic potential at a point ( (0, 0, z) ) due to a dipole-like configuration of charges ( q ) and ( -q ) at ( (0, 0, -a) ) and ( (0, 0, a) ) respectively can be calculated using the derived formula. The potential at ( (0, 0, 0) ) is zero.

Keywords

Electrostatic Potential, Charge Distribution, Electric Field