Coulombs Law with vectors question help, test tomorrow.

[skg94]

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

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 \sqrt{235..^2 + 176...^2} 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

 

[ehild]

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

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^2

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 \sqrt{235..^2 + 176...^2} 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

You made a mistake when calculating the hypotenuse: .015625 is the square of of the hypotenuse, you do not need to square again in the equation for the force. .

ehild