Question #178422

Three point charges are placed at the following points on the x-axis: +2 μC at x = 0, - 3 μC at x = 40 cm, - 5μC at x = 120 cm. 

  1. Find the force on the - 3 μC charge
  2. Find the force on the - 5 μC charge
1
Expert's answer
2021-04-06T13:53:28-0400

(a) Let's find the net electric force on the -3.0μC charge:


Fnet=F12F32,F_{net}=F_{12}-F_{32},Fnet=k(q1q2(r12)2q3q2(r13r12)2),F_{net}=k(\dfrac{|q_1q_2|}{(r_{12})^2}-\dfrac{|q_3q_2|}{(r_{13}-r{_{12}})^2}),

Fnet=9109 Nm2C2(2106 C(3106 C)(0.4 m)2(5106 C)(3106 C)(1.2 m0.4 m)2)=0.13 N.F_{net}=9\cdot10^9\ \dfrac{Nm^2}{C^2}\cdot(\dfrac{|2\cdot10^{-6}\ C\cdot(-3\cdot10^{-6}\ C)|}{(0.4\ m)^2}-\dfrac{|(-5\cdot10^{-6}\ C)\cdot(-3\cdot10^{-6}\ C)|}{(1.2\ m-0.4\ m)^2})=0.13\ N.

(b) Let's find the net electric force on the -5.0μC charge:


Fnet=F13+F23,F_{net}=F_{13}+F_{23},Fnet=k(q1q3(r13)2+q2q3(r13r12)2),F_{net}=k(\dfrac{|q_1q_3|}{(r_{13})^2}+\dfrac{|q_2q_3|}{(r_{13}-r_{12})^2}),

Fnet=9109 Nm2C2(2106 C(5106 C)(1.2 m)2+(3106 C)(5106 C)(1.2 m0.4 m)2)=3.041011 N.F_{net}=9\cdot10^9\ \dfrac{Nm^2}{C^2}\cdot(\dfrac{|2\cdot10^{-6}\ C\cdot(-5\cdot10^{-6}\ C)|}{(1.2\ m)^2}+\dfrac{|(-3\cdot10^{-6}\ C)\cdot(-5\cdot10^{-6}\ C)|}{(1.2\ m-0.4\ m)^2})=3.04\cdot10^{-11}\ N.

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