Two charges are fixed in place with a separation d. One charge is positive and has nine times (n = 9) the magnitude of the other charge, which is negative. The positive charge lies to the left of the negative charge. Relative to the negative charge, locate the two spots on the line through the charges where the total potential is zero.
positive distance to the left of -q =?
positive distance to the right of -q =?
Potential is additive. Put the + charge at x=0.
Now, test the potential at x
V= k9q/x - kq/(d-x)
Set v to zero, and solve for x.
To find the positions on the line where the total potential is zero, we need to solve for the values of x, which represent the distances from the negative charge.
First, let's set up the expression for the total electric potential:
V = k * 9q / x - k * q / (d - x)
Where:
- V is the total potential
- k is the Coulomb's constant
- q is the magnitude of the negative charge
- d is the separation between the charges
- x represents the distance from the negative charge
We want to find the values of x where V is equal to zero. So, let's set V equal to zero:
0 = k * 9q / x - k * q / (d - x)
Now, we can simplify the equation by multiplying both sides by x(d - x) to eliminate the denominators:
0 = 9q * (d - x) - qx
Simplify further:
0 = 9qd - 9qx - qx
0 = 9qd - 10qx
Now, let's solve for x by isolating it:
10qx = 9qd
x = (9qd) / (10q)
Cancel out the q:
x = (9d) / 10
So, the positive charge is located at a distance of (9d)/10 to the left of the negative charge. And since their separation is d, the positive charge will be at a distance of d - (9d)/10, which simplifies to d/10, to the right of the negative charge.
Therefore, the positive distance to the left of -q is (9d)/10, and the positive distance to the right of -q is d/10.