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skg94
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1. Homework Statement [/b]
1. Charge A (-2uC) is 0.10m left of charge B (+3uC), with charge C (+4uC), 0.075m below charge B, forming a right angle triangle with the right angle at B. Find the net electrostatic charge on C
if its confusing the 0.10 is the opposite of the hypotenuse with the 0.075m being the adjacent.
i don't how to draw a damn triangle i can't do it with just normal symbols but if oyu don't understand the triangle i will try to explain it further
Fe=kq1q2/r^2
Fe(b on c) = (8.99*10^9)(3*10^-6)(4*10^-6)/ .075^2 = 19.17866666..
Fe (a on c) = first i found the hypotenuse, (.1^2+.075^2) = .015625m
then found Fe, K * ( 4*10^-6) (2*10^-6)/.015625^2 = 294.58432
I found the angle at C, by using pythagoras, tan-1(0.10/.075) = 53.13010235, then found the angle to use to find the x and y component of Fe (a on c) by subtracintg 53.. from 90, which is 36.86089765..
Fe (aonc x) = 294... * cos36... = 235.667456
Fe (aonc y) = 294... * sin 36... = 176.750592
then [itex]\sqrt{235..^2 + 176...^2}[/itex] which equals back to 294.58432.
I may have done the whole Fe ( a on c) wrong, if i did i don't know where.
The answer is 16.8N 12.6 degrees W of S
1. Charge A (-2uC) is 0.10m left of charge B (+3uC), with charge C (+4uC), 0.075m below charge B, forming a right angle triangle with the right angle at B. Find the net electrostatic charge on C
if its confusing the 0.10 is the opposite of the hypotenuse with the 0.075m being the adjacent.
i don't how to draw a damn triangle i can't do it with just normal symbols but if oyu don't understand the triangle i will try to explain it further
Homework Equations
Fe=kq1q2/r^2
The Attempt at a Solution
Fe(b on c) = (8.99*10^9)(3*10^-6)(4*10^-6)/ .075^2 = 19.17866666..
Fe (a on c) = first i found the hypotenuse, (.1^2+.075^2) = .015625m
then found Fe, K * ( 4*10^-6) (2*10^-6)/.015625^2 = 294.58432
I found the angle at C, by using pythagoras, tan-1(0.10/.075) = 53.13010235, then found the angle to use to find the x and y component of Fe (a on c) by subtracintg 53.. from 90, which is 36.86089765..
Fe (aonc x) = 294... * cos36... = 235.667456
Fe (aonc y) = 294... * sin 36... = 176.750592
then [itex]\sqrt{235..^2 + 176...^2}[/itex] which equals back to 294.58432.
I may have done the whole Fe ( a on c) wrong, if i did i don't know where.
The answer is 16.8N 12.6 degrees W of S
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