Question

One electron is at the origin of a coordinate system. A second electron, somewhere in the x−y plane, feels a force 2.2×10−22Ni^+3.6×10−22Nj^.

Where is the second electron located?

Express your answers using two significant figures. Enter the x and y coordinates of the second electron separated by a comma.

I keep getting 1.02 x 10^-3 m, 7.99 x 10^-4 m but I put it in as 8.0 x 1-^-4 and it is saying its wrong

Answer 1

et the electron at the origin be e1
Let the second electron be e2
The information we have been given allows to work out an equation for the strength of the E-field from the e1. Once we have that, we can use Coulomb's law to get an equation that will give us straight line distance from e1 to e2, and its direction from the origin.

We are told that the force on e2 is
F(e2) = (2.0 * 10^-22) i + (3.4 * 10^-22) j

Notice that the components of the force are both positive. So e2 must be moving left to right and upwards. We know that electrons repell each other, so e2 can't be in the second or third quadrants, otherwise it would be moving towards e1 instead of away from it.

If it was in the fourth qudrant, it would be moving in the x-direction but also in the y-direction, towards e1. The only quadrant it can be in is the first.
Now we have established that, we can get an equation for the strength of the E-field of e1.

Recall that

Magnitude of E-field = Force / charge

E = F/ -e1
But the E-field of an electron points inwards, so will be negative.

- E = F / -e1
the charge on an electron is - 1.6 * 10^-19C

- E = (2 .2* 10^-22)/ (- 1.6 * 10^-19) i + (3.6 * 10^-22 / -1.6 * 10^-19) j

- E = - (1.37 ^ 10^-3) i - (2.25 * 10^-3) j

Multiply through by minus 1

E = (1.37 ^ 10^-3) i + (2.25 * 10^-3) j

From this, using Pythagoras, we can now work out the magnitude and direction of E

Magnitude = √[(1.37 * 10^-3)^2 + (2.25 * 10^-3)^2]

IEI = √ [(1.876 * 10^-6) + (5.06 * 10^-6)]
IEI = √(6.08 * 10^-6)
IEI = 2.632 ^ 10^-3 N/C

From Coulomb's law

E = Ke1/r^2

r^2 = Ke1 / E
r = √(Ke1/r^2)

r = √ (9 * 10^9 * 1.6 * 10^-19/ 2.63 ^ 10^-3)

r = √ (5.47 * 10^-7)
r = √(0.547 * 10^-6)
r = 7.39 * 10^-3 m
This is the straight line distance from e1 to e2.

From basic trig, its direction is
arctan (1.37/2.25) = 31.33°