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

In summary: So we only need to solve the first two cases.k(-2)/x + k(3)/|L-x| = 0Multiply both sides by x(L-x)k(-2)(L-x) + k(3)x = 0Simplify and rearrange to get a quadratic equation in x:2kx^2 - 5kLx + 6kL = 0Solve for x using the quadratic formula:x = (5kL ± √(25k^2L^2 - 48k^2L)) / 4k = (5L ± √(25L^2 - 48L)) / 4= (5L
  • #1
dmitriylm
39
2

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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  • #2
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
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
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
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
dmitriylm said:
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
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
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
smoics said:
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.
 

1. 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 denoted by the symbol V and is measured in volts (V).

2. How is electric potential calculated?

Electric potential is calculated using the equation V = kQ/r, where V is the electric potential, k is the Coulomb's constant, Q is the charge at the point, and r is the distance between the point and the source of the electric field.

3. Why is the electric potential zero at certain points on the x-axis?

The electric potential is zero at certain points on the x-axis because the electric field is perpendicular to the x-axis at those points. This means that no work is done in moving a charge along the x-axis, resulting in zero change in potential energy and therefore, zero electric potential.

4. Can the electric potential ever be negative?

Yes, the electric potential can be negative. This occurs when the charge at the point is negative or when the electric field is directed opposite to the direction of motion of the charge. The negative sign indicates that work must be done to move a charge against the electric field, resulting in a decrease in potential energy and a negative electric potential.

5. How does the electric potential change as we move along the x-axis?

The electric potential generally decreases as we move along the x-axis, as the distance between the point and the source of the electric field increases. However, there may be points along the x-axis where the electric potential is zero or increases, depending on the distribution and strength of the electric field at those points.

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