At which points on the x-axis (not at infinity) is the electric potential zero?

  • Thread starter dmitriylm
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  • #1
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Homework Statement



Two point charges -2Q and +3Q are on the x-axis at the origin and at
x = L. Find all the points on the x-axis (not at infinity) where the electric
potential is zero. Express your answers in terms of L and Q.


Homework Equations



V= k*Q/r


The Attempt at a Solution



I understand that when dealing with two positive charges there is a point between them (if they are equal magnitude, this would be right at d/2) where the field magnitude is zero (because the fields repel each other) but with two opposite charges am I wrong to assume that the only place where the potential would be zero would be directly at the negative charge as field lines always point in the direction of decreasing potential? How would I express this?
 
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Answers and Replies

  • #2
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The electric field decreases from the point proportional to 1/r^2 so at distances close to one of the charges it's electric field dominates. So you are looking for a point closer to your -Q point since it is a lower magnitude.
 
  • #3
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That's what I was thinking, that zero potential would be directly at the origin. How would I express this in terms of L and Q?
 
  • #4
rl.bhat
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The potentials will be zero at the points where pogtential due to -2Q and 3Q are equal. You will get two such points. One in between the charges and another out side the two charges, nearer to the smaller charge.
 
  • #5
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Would this be correct:

k(-2)/L + k(+3)/L = 0

-2/L + 3/L = 0

-2/L = -3/L

Potential is equal to zero at L = -2 and -3?
 
  • #6
rl.bhat
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Would this be correct:

k(-2)/L + k(+3)/L = 0

-2/L + 3/L = 0

-2/L = -3/L

Potential is equal to zero at L = -2 and -3?
N0. Let x be the distance from -2 where the potential is zero. So
k(-2)/(x) = k(+3)/(L-x)
For the second answer
k(-2)/(x) = k(+3)/(L+x)
Solve for x.
 
  • #7
SammyS
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If the ‒2Q charge is at the origin and the 3Q charge is at x=L, then the electric potential at location x on the x-axis is given by:

[tex]V=k\frac{-2Q}{|x|}+k\frac{3Q}{|L-x|}[/tex]

Set V=0 and solve for x.

BTW: You are looking for the location(s) where electric potential is zero, NOT the location(s) where the electric field is zero.
 
  • #8
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How do you end up with more than one answer for x, using the above location? Do you have to use the abs. value of L-x or something??
 
  • #9
SammyS
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How do you end up with more than one answer for x, using the above location? Do you have to use the abs. value of L-x or something??
Yes, you need to use abs. value of both x and L-x.

0 < x < L : If x is between 0 & L, assuming L is a positive number, then |x| = x, and |L-x| = L-x.

x < 0 < L: If x is negative, then |x| = -x, and |L-x| = L-x

0 < L < x : If x is greater than L, then |x| = x, and |L-x| = -(L-x) = x-L

Since we're looking for V=0, the last two cases give equivalent equations.
 

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