Consider two charges, where one (+1.90 nC) is at the origin and the other (-3.61 nC) is at the position x = 4.35 cm.
Find the x-coordinate where a proton would experience zero net force.
here.... +small ...... -big
x ......... 0.......... 4.35
just using distances without regard to sign to get equal field magnitudes
1.90/x^2 = 3.61/(x+4.35)^2
solve for x, note it has to be - (left of origin)
3.61 x^2 = 1.9(x^2 +8.7 x+18.9)
3.61x^2 = 1.9x^2 + 16.5x + 35.9
1.71x^2 -16.5x -35.9 = 0
|x| = either 11.4 or 1.83
try -1.83
check
.567 = 1.9/(1.83)^2
3.61/(6.18)^2 = .09 nope
try -11.4
.016 = 1.9/11.4^2
3.61/(225) = .016
So x = -11.4
To find the x-coordinate where a proton would experience zero net force, we need to calculate the electric field created by each charge and determine the point where the electric field due to the positive charge cancels out the electric field due to the negative charge.
First, let's calculate the electric field at a point x due to the positive charge at the origin. The electric field due to a point charge is given by the formula:
E = k * Q / r^2
Where:
- E is the electric field
- k is Coulomb's constant (k = 9.0 x 10^9 N m^2 / C^2)
- Q is the charge
- r is the distance from the charge to the point
In this case, Q = +1.90 nC (or +1.90 x 10^-9 C) and r = x.
Therefore, the electric field at point x due to the positive charge is:
E1 = k * Q1 / r^2
= (9.0 x 10^9 N m^2 / C^2) * (1.90 x 10^-9 C) / x^2
Now, let's calculate the electric field at point x due to the negative charge at x = 4.35 cm. Similarly, we have:
Q2 = -3.61 nC (or -3.61 x 10^-9 C)
r = x - 4.35 cm (converting cm to meters by dividing by 100)
Therefore, the electric field at point x due to the negative charge is:
E2 = k * Q2 / r^2
= (9.0 x 10^9 N m^2 / C^2) * (-3.61 x 10^-9 C) / (x - 4.35 x 10^-2 m)^2
To find the x-coordinate where the net force on a proton is zero, the electric field due to the positive charge and the electric field due to the negative charge must be equal in magnitude but have opposite directions. So, we set E1 = -E2 and solve for x.
(9.0 x 10^9 N m^2 / C^2) * (1.90 x 10^-9 C) / x^2 = (9.0 x 10^9 N m^2 / C^2) * (-3.61 x 10^-9 C) / (x - 4.35 x 10^-2 m)^2
By solving this equation for x, we can find the x-coordinate where a proton would experience zero net force.