Question

Charges of 4.0

Answer 1

Hi there, I'm just studying for my physics 102 exam thats on saturday and was trying to work out this problem also. It is kind of hard but i think that i have figured it out. This is what I did...

1. first find out the Electric Field (E) on each charge by the equation E=k(q)/r^2, k is a constant = 8.99 x 10^9 C. So once you do that for each charge you should get that E for the 4 x 10^6 C is equal to 3.596 x 10^6, and for the -6 x 10^6C equals -5.394 x 10^6.
2. To find out the E on the third corner you have to take into account the x and y of E on the third corner, so you can plug it into the formula Etot = square root of Ex^2 + Ey^2. For Ex, you use the formula Ex = [k(q1-q2)/r^2] x cos60. watch the signs, you will end up having (8.99 x 10^9(4 x 10^-6 + 6 x 10^-6)/(.1^2) **because subtracting a negative is equal to just plain addition. For Ey=[k(q1 + q2)/r^2] x sin60. So hopefully you'll end up with Ex=4.5 x 10^6, Ey=1.6 x 10^6.
3. Finally you use the formula previously stated, being Etotal=square root of Ex^2 + Ey^2. So simply you enter your Ex and Ey values such as Etot=square root of (4.5 x 10^6)^2 + (-1.6 x 10^6)^2 * watch that when you square a negative number it now becomes positive. So youll have Etot=square root of (2.025 x 10^13) + (2.56 x 10^12). Add those together and square root that number and you should get the final answering being...4.8 x 10^6.
Hopefully you understood all of that, and it helps you out