Charges of −q and +2q are fixed in place, with a distance of 4.65 m between them. A dashed line is drawn through the negative charge, perpendicular to the line between the charges. On the dashed line, at a distance L from the negative charge, there is at least one spot where the total potential is zero. Find L.
potential due to -q = -kq/L
potential due to 2Q = +2k q/(L^2+4.65^2)^.5
so if
1/L = 2/(L^2+4.65^2)^.5
2 L = sqrt(L^2 + 4.65^2)
4 L^2 = L^2 + 4.65^2
3 L^2 = 21.6
L = 2.68
To find the distance L at which the total potential is zero on the dashed line, we need to consider the electric potential due to both charges at that point.
The electric potential at a point due to a point charge q can be calculated using the formula:
V = k * q / r
Where V is the electric potential, k is the electrostatic constant (k = 8.99 x 10^9 Nm^2/C^2), q is the charge, and r is the distance from the charge to the point.
Let's denote the distance between the negative charge (-q) and the point as x, and the distance between the positive charge (+2q) and the point as (4.65 - x). Since the dashed line is perpendicular to the line between the charges, we can draw a right-angled triangle to represent the distances.
Using the formula mentioned above, the electric potentials due to both charges can be calculated as follows:
Negative charge potential (V1) = k * (-q) / x
Positive charge potential (V2) = k * (+2q) / (4.65 - x)
Since we want the total potential to be zero, we can set V1 + V2 = 0:
k * (-q) / x + k * (+2q) / (4.65 - x) = 0
Simplifying the equation:
(-q) / x + 2q / (4.65 - x) = 0
Next, we can multiply through by x(4.65 - x) to get rid of the denominators:
(-q)(4.65 - x) + 2qx = 0
Expanding and rearranging the equation:
4.65q - qx - 2qx = 0
4.65q - 3qx = 0
Bringing the terms involving x to one side:
3qx = 4.65q
Simplifying:
qx = (4.65q) / 3
x = (4.65q) / (3q)
Canceling out the q terms:
x = 4.65 / 3
Therefore, the distance L from the negative charge where the total potential is zero on the dashed line is L = 4.65 / 3.