Electric potential due to two point charges near origin

In summary, the potential V at a point (x,y,z) near to the origin, due to two point charges (+q) fixed at (±a,0,0), can be expressed as V = [q/4∏*ε*a] [[2+(2x2-y2-z2)/a2], where x, y and z are small compared with a. This is derived using the equations V(x) = q/4∏i*ε*r and superposition, as well as the binomial expansion for small x. The final answer should keep terms up to the second order.
  • #1
raving_lunatic
21
0

Homework Statement


A charge + q is fixed at each of the two points (±a,0,0). Show that the potential V at a point (x,y,z) near to the origin may be expressed as

V = [q/4∏*ε*a] [[2+(2x2-y2-z2)/a2]
where x, y and z are small compared with a.


Homework Equations



V(x) = q/4∏i*ε*r , where r is the distance between the point charge and x and V is the potential at x.

Superposition, i.e. V due to multiple point charges = the sum of V due to each point charge

(1+x)^n = 1+nx+... valid for small x.

The Attempt at a Solution



This seemed like an incredibly straightforward problem, which is probably why it's quite frustrating. I tried adding together the potentials due to each point charge, approximated using the binomial expansion:

V(q1) = q/4∏ε[(x+a)2 + y2 +z2]^-1/2

V(q2) = q/4∏εa[(a-x)2 + y2 +z2]^-1/2

Taking a out of the brackets as a common factor gives:

V(q1) = q/4∏εa [(x/a + 1)2+ (y2+z2)/a2)]^-1/2

and a similar expression for V(q2) but with (1-x/a)2.

I then expanded the squares and used (2x/a + (x2+y2+z2)/a2) as a small parameter to perform the binomial expansion; this gave

V(q1) = q/4∏εa [1 - x/a - (x2+y2+z2/2a2)]

V(q2) = q/4∏εa [1 + x/a - (x2+y2+z2/2a2)]

Which doesn't lead to the desired result when added; where did I go wrong?
 

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  • #2
I think I've worked it out; need to keep the terms to the second order. Sorry.
 

What is electric potential?

Electric potential is the amount of electric potential energy per unit charge at a given point in an electric field. It is a measure of the work required to move a unit charge from infinity to that point in the electric field.

How do you calculate electric potential due to two point charges near the origin?

The electric potential due to two point charges near the origin can be calculated using the formula V = k(Q1/r1 + Q2/r2), where V is the electric potential, k is the Coulomb's constant, Q1 and Q2 are the charges of the two point charges, and r1 and r2 are the distances from the origin to each point charge, respectively.

What is the unit of electric potential?

The unit of electric potential is volt (V), which is equivalent to joules per coulomb (J/C).

How does the distance between two point charges affect the electric potential?

The electric potential between two point charges is inversely proportional to the distance between them. As the distance increases, the electric potential decreases and vice versa.

What is the difference between electric potential and electric potential energy?

Electric potential is a measure of the electric potential energy per unit charge at a given point in an electric field, while electric potential energy is the energy that a charged particle possesses due to its position in an electric field. In other words, electric potential is a property of the electric field, while electric potential energy is a property of the charged particle.

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