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## Electric Potential  3-Visitor

## Electric Potential

On 9/5/2003 12:14:35 PM, abk511 wrote:

>where
>r1=Squareroot[(x+R)^2+y^2+z^2]
>and
>r2=Squareroot(x^2+y^2+z^2) I
>don't understand why r1 and r2
>me more details. Thanks

Like the equation for a circle, the equation for a sphere has radius=sqrt[(x-a)^2+(y-b)^2+(z-c)^2]

TTFN,
Eden
3 REPLIES 3  3-Visitor
(To:IRstuff)
one charge is located at -R and the other is located at 0

TTFN,
Eden  3-Visitor
(To:IRstuff)
Go back to the original equation for the potential:
V(x,y,z)= ke(q)/r1+ke(-2q)/r2

We're setting V(x,y,z)=0 and we get an equation that, potentially ;-), could be the equation for a sphere, by substituting the equations for r1 and r2 into the potential equation.

Note that k and q can be divided out, move the 2/r2 term to the other side of the equal side, squaring both sides, leaving

1/r12=4/r22, now mutliply both sides by r12r22, resulting in:

r22=4r12, now you should be able to complete the square and get an equation for a sphere.

TTFN,
Eden  3-Visitor
(To:IRstuff)
x2+y2+z2=4[(x+R)2+y2+z2]

subtract the left side from both sides:

0=3x2+8xR+4R2+3y2+3z2

divide out the 3
0=x2+8/3xR+4/3R2+y2+z2

to complete the square for the x-term, we need 16/9R2, but we only have 12/9R2, so we need to add 4/9R2 to both sides.

Can you take it from here?

TTFN,
Eden 