- #1
Genericcoder
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Hi guys I came upon this problem while solving in a book, there is something I am confused about the way he solved it in the answers.
(II)Three charged particles are placed at the corners of an
equilateral triangle of side 1.2m. The charges are 4uC,-8uC,and
-6uC. Calculate the magnitude and direction of the net force
on each due to the other two.
The charge -4uC is on top and -6.00uC is on the right and -8uC is on the left
I answered this by calculating the magnitude of each charge then getting the net charge by adding the vector components of each.
F12 = 0.2N;
F13 = 0.15N;
F32 = 0.3N;
F31 = 0.15;
F23 = 0.3N;
F21 = 0.2N;
I concluded the force on charge Q1 should be pointing to Q2 since its larger magnitude so I first calculated the components vectors F12 as negative as so:-
F12x = -cos(60)*0.2 = -0.1N;
F12y = -sin(60) * 0.2) = (-(3^1/2) / 10) N
F13x = cos(60)(0.15) = (3/40)N
F13y = sin(60)(0.15) = 0.129N
Now since the force is favored to go to F12 then we can add F12 to F13 we are subtracting anyway that's why I did F13 as positive,since I know that force net should go to F12;
F1x = -0.3 + 3/40 = -1/40N
F1y = -(3^1/2) / 10 + 0.129 = -0.044N
But the way he answered in the book is really different.
the book answer is as follows:
F1x = -(0.1997N)cos(60) + (0.1498)cos60 = -2.495 * 10^-2;
F1y = -(0.1997)sin(60) - (0.1498N)sin60 = -3.027 * 10^-1;
Notice that he subtracted the y component vector which really doesn't make sense I mean both are attracting the positive proton so it wouldn't make sense to subtract. Did I miss something here? I also notice that 0.1997 F12 is really close just my rounding must have been incorrect. I also didn't include here the calculation for the force relative to point 2,since if I know what I did wrong here same thing apply to point 2 calculation(thats if I did something wrong).
Thanks alot.
(II)Three charged particles are placed at the corners of an
equilateral triangle of side 1.2m. The charges are 4uC,-8uC,and
-6uC. Calculate the magnitude and direction of the net force
on each due to the other two.
The charge -4uC is on top and -6.00uC is on the right and -8uC is on the left
I answered this by calculating the magnitude of each charge then getting the net charge by adding the vector components of each.
F12 = 0.2N;
F13 = 0.15N;
F32 = 0.3N;
F31 = 0.15;
F23 = 0.3N;
F21 = 0.2N;
I concluded the force on charge Q1 should be pointing to Q2 since its larger magnitude so I first calculated the components vectors F12 as negative as so:-
F12x = -cos(60)*0.2 = -0.1N;
F12y = -sin(60) * 0.2) = (-(3^1/2) / 10) N
F13x = cos(60)(0.15) = (3/40)N
F13y = sin(60)(0.15) = 0.129N
Now since the force is favored to go to F12 then we can add F12 to F13 we are subtracting anyway that's why I did F13 as positive,since I know that force net should go to F12;
F1x = -0.3 + 3/40 = -1/40N
F1y = -(3^1/2) / 10 + 0.129 = -0.044N
But the way he answered in the book is really different.
the book answer is as follows:
F1x = -(0.1997N)cos(60) + (0.1498)cos60 = -2.495 * 10^-2;
F1y = -(0.1997)sin(60) - (0.1498N)sin60 = -3.027 * 10^-1;
Notice that he subtracted the y component vector which really doesn't make sense I mean both are attracting the positive proton so it wouldn't make sense to subtract. Did I miss something here? I also notice that 0.1997 F12 is really close just my rounding must have been incorrect. I also didn't include here the calculation for the force relative to point 2,since if I know what I did wrong here same thing apply to point 2 calculation(thats if I did something wrong).
Thanks alot.