Question

three point charges are arranged along the x-axis. charge q1= -4.50 nC is located at x =0.200 m, and charge q2= +2.50 nC is at x= -0.300m. a positive point charge q3 is located at the origin. (a) what must the value of q3 be for the net force on this point charge to have magnitude 4.00 un? (b) what is the direction of the net force on q3? (c) where along the x axis can q3 be placed and the net force on the zero, other than the trivial answers of x= +infinity and x= -infinity

Answer 1

$F_{net}=\frac {kq_1q_2}{r^2}+\frac {kq_2q_3}{r^2}+\frac {kq_1q_3}{r^2}$

Put value

$F_{net}=\frac {k(-4.50\times10^{-9})(2.50\times10^{-9})}{(5\times10^{-1})^2}+\frac {k2.50\times10^{-9}q_3}{(3\times10^{-1})^2}+\frac {k4.50\times10^{-9}q_3}{(2\times10^{-1})^2}$

$F_{net}=4\times10^{-6}N$

Part (1)and(2) is equal

$4\times10^{-6}=\frac {k(-4.50\times10^{-9})(2.50\times10^{-9})}{(5\times10^{-1})^2}+\frac {k2.50\times10^{-9}q_3}{(3\times10^{-1})^2}+\frac {k(-4.50)\times10^{-9}q_3}{(2\times10^{-1})^2}$ $q_3=-5.77nc$

(b)net forces direction along to x- axis to the left

(c) out side in line point p persent

$F_{net}=\frac{-4.50\times q_3}{x^2}+\frac{2.50q_3}{(x-5)^2}=0$

$9(x-0.5)^2=5x^2$

$x=11.208meter$

For -4.50 charge

X=11.208meter