Question

Two point charges are placed as follows: charge q_{1} = - 1.50nC is at y = 6.00 m and charge q_{2} = 3.20nC is at the origin. What is the total force (magnitude and direction) exerted by these two charges on a negative point charge q_{3} = - 5.00nC located at (2.00 m, -4.00 m)?

Answer 1

Solution.

$q1=-1.50\sdot10^{-9} C;$

$q2=3.20\sdot10^{-9} C;$

$y=6.00 m;$

$q3=-5.00\sdot10^{-9}C; (2.00m;-4.00m);$

$F-?;$

$F_{13}=k\dfrac{q_1q_3}{r_{13}^2};$

$F_{13}=9\sdot10^9\dfrac{1.5\sdot10^{-9}\sdot5.00\sdot10^{-9}}{10^2+2^2}=6.49\sdot10^{-10}N;$

$F_{23}=k\dfrac{q_2q_3}{r_{23}^2};$

$F_{23}=9\sdot10^9\dfrac{3.2\sdot10^{-9}\sdot5.00\sdot10^{-9}}{4^2+2^2}=72\sdot10^{-10}N;$

$F_{13x}=F_{13}\sdot cos\alpha=6.49\sdot10^{-10}\sdot\dfrac{2}{\sqrt{10^2+2^2}}=1.27\sdot10^{-10}N;$

$F_{13y}=F_{13}\sdot sin\alpha=6.49\sdot10^{-10}\sdot\dfrac{10}{\sqrt{10^2+2^2}}=6.36\sdot10^{-10}N;$

$F_{23x}=F_{23}\sdot cos\alpha=72\sdot10^{-10}\sdot\dfrac{2}{\sqrt{4^2+2^2}}=32.20\sdot10^{-10}N;$

$F_{23y}=F_{23}\sdot sin\alpha=72\sdot10^{-10}\sdot\dfrac{4}{\sqrt{10^2+2^2}}=64.40\sdot10^{-10}N;$

$F_x=F_{13x}-F_{23x}=1.27\sdot10^{-10}-32.20\sdot10^{-10}=-30.93\sdot10^{-10}N;$

$F_{y}=F_{23y}-F_{13y}=64.40\sdot10^{-10}-6.36\sdot10^{-10}=58.04\sdot10^{-10}N;$

$F=\sqrt{F_{x}^2+F_{y}^2}=\sqrt{(-30.93\sdot10^{-10})^2+(58.04\sdot10^{-10})^2}=65.80\sdot10^{-10}N;$

$tan\gamma=\dfrac{58.04\sdot10^{-10}}{30.93\sdot10^{-10}}=1.876;$

$\gamma=62^o or118^o;$

Answer: $F=65.80\sdot 10^{-10}N, \gamma=62^o or 118^o.$