Question

A charge of -3.00 nC is placed at the origin of an x-y coordinate system, and a charge of 2.00 nC is placed on the y-axis at y = 4.00 cmb.

If a third charge of 5.00 nC is now placed at the point x = 3.00 cm, y = 4.00 cm, what are the x and y components of the total force exerted on the charge by the other two charges?c.e.

What are the magnitude and direction of the total force?

I would appreciate an explanation of the magnitude and force part , thank you so much !

Answer 1

for calculation convenience let me take the charges to be 5 units, -2 units and 6 units. instead of 3 ,2 and 5.
Let P be the point where the third charge is placed.
The distances for 6 C charge is 3 units from -2 C and 5 units right from the origin. (3^2 + 4^2 = 5^2)
So force F1 due to 5C on 6C will be K * 30/25 along the OP repulsive along OP
The force F2 due to -2 C on the same 6 C will be K * 12/9 attractive acting along PM
M is the position of -2 C on x axis.
Now x component of repulsive is F1 cos@ where tan @ = 3/4
And x component of attractive is 0
So total can be found out
Now y component of F1 is F1 sin@ and that of F2 is -ve
So we have to get the difference of these two
Now knowing total of x and y components you can find the net by ./ [(x comp)^2 + (ycomp)^2]
./ is the Square root symbol
To get the direction tan m = y comp total / x comp total
Right from this required angle between F and x axis can be computed.