Two negative and two positive point charges (magnitude Q = 3.86mC ) are placed o
ID: 2233866 • Letter: T
Question
Two negative and two positive point charges (magnitude Q = 3.86mC ) are placed on opposite corners of a square as shown in the figure (a= 0.145m) Determine the magnitude of the force on each charge.
I need someone to look at my work and tell me whereI went wrong...First, I understand that force needs to be separated into X vs Y components.
F13 is separated into X and Y components, named F13x and F13y:
The distance between 1 and 3 is the diagonal of the square...we use pythagorean to get that.
c^2 = a^2 + b^2, since it's square...0.145^2 + 0.145^2
c = 0.205...
F3 = kQ1Q3 / c^2 = (9E9)(3.86E-6)^2 / 0.205^2
F3 = 3.189
This gives us F3, which is the hypotenuse. F13x and F13y can be found using this number
F13x = (3.189)(cos45) = 2.254. We call this negative ecause the vetor points AWAY from the chrge Q1.
F13y = (3189)(sin45) = 2.254
For TOTAL X Components:
F12 (force of 2 on 1) and F13x (the X compo of 3 on 1):
F12 = kQ1Q2/a^2 = (9E9)(3.86 E-6)^2 / (0.145^2) = 6.377
F13x = -2.254
Fx total = F12 + F13x = 6.377 + -2.254= 4.122
For TOTAL Y components:
F14 = -6.377 (it points downward in negative y axis direction)
F13y = 2.254
Fytotal = -4.122
This makes another triangle...where F13y points up and 13x points away from the Q1 charge.
pythagorean theorem again: (-4.122)^2 +(4.122)^2 = F1
F1 = 5.8307 N
And Mastering says it's wrong and I don't know what I'm doing wrong at all.
Ugh, someone please help. I'm sorry couldn't draw al the triangles.
Explanation / Answer
in the question it is given
Two negative and two positive point charges (magnitude Q = 3.86mC ) are ....
=> Q=3.86E-3
but for calculations you have used Q=3.86E-6
I have checked all calculation they are perfect and so is the procedure
so the answer should be multiplied with factor 1E6 to get correct results
F1 = 5.8307E6 N
and derection will be the same
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