# How to Show Derivatives of 1/R for a 1D Charge Distribution?

• raintrek
In summary, the question is asking for methods to solve for the derivatives of 1/R and 1/R^2 in a 1 dimensional problem, using the given information about charges and their positions along the x-axis. The solution involves substituting R with x and using basic derivative rules.
raintrek

## Homework Statement

"Assume that charges +q, -3q, 2q lie at positions -2a, 0, +2a along the x-axis respectively.")

I've calculated dipole/quadrupole moments about the origin as well as the exact potential at x=+10a, however I'm confused by this next part to the question:

Show that, for a 1 dimensional problem,

d(1/R) / dx = -1/x²

and

d²(1/R) / dx² = 2/x³

where R is the distance between a lab fixed point x and an arbitrary origin x0.

Can anyone suggest a method of tackling this??

raintrek said:
Show that, for a 1 dimensional problem,

d(1/R) / dx = -1/x²

and

d²(1/R) / dx² = 2/x³

where R is the distance between a lab fixed point x and an arbitrary origin x0.

Can anyone suggest a method of tackling this??

I am at a loss to understand what this has to do with charges.

R has been defined to be equal to x.

So, d/dx(1/R) = d/dx(1/x) = -1/x^2 etc.

## What is a 1 dimensional charge distribution?

A 1 dimensional charge distribution refers to a system in which charges are distributed along a single line or axis. This is in contrast to a 2 or 3 dimensional charge distribution where charges are distributed in a plane or in space, respectively.

## How is a 1 dimensional charge distribution different from a point charge?

A point charge is a theoretical concept in which all of the charge is concentrated at a single point. In a 1 dimensional charge distribution, the charge is spread out along a line, resulting in a more realistic and complex distribution of charge.

## What is the electric field produced by a 1 dimensional charge distribution?

The electric field produced by a 1 dimensional charge distribution depends on the distribution of charges and their magnitudes. This can be calculated using the principle of superposition, where the electric field due to each individual charge is added together.

## Can a 1 dimensional charge distribution create a uniform electric field?

No, a 1 dimensional charge distribution cannot create a uniform electric field. This is because the electric field produced by each individual charge decreases with distance, resulting in a non-uniform field overall.

## What are some real-life examples of 1 dimensional charge distributions?

Some examples of 1 dimensional charge distributions include a long, thin wire with a uniform charge distribution, a line of evenly spaced electrically charged particles, or a charged cylinder with a finite length.

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