Answer to Question #121697 in Electricity and Magnetism for Maximo

Question

A positive charge Q1 = +250.0 μC is located at a point X1 = -2.00 m, a negative charge Q2 = -150.0 μC is located at a point X2 = 3.00 m and a negative charge Q3 = -200.0 μC is located at a point X3 = 6.00 m.a. Draw free body diagrams for the electric force acting on Q1, Q2 and Q3.b. Find the magnitude of the force between Q1 and Q2.c. Find the magnitude of the force between Q1 and Q3.d. Find the magnitude of the force between Q2 and Q3.e. Find the magnitude and direction of the net electric force on charge Q1.f. Find the magnitude and direction of the net electric force on charge Q2.g. Find the magnitude and direction of the net electric force on charge Q3.

Answer 1

Let,

$$k=\frac{1}{4\pi\epsilon_o}=9\times 10^9$$

We know that

$$F_{12}=k\frac{q_1q_2}{d^2_{12}}$$

(a).

b).

$$Q_1=250\mu C,Q_2=-150\mu C,d_{12}=5m\\ \implies F_{12}=k\frac{250\times 150\times 10^{-12}}{25}=15\times 10^{-10}k=13.5N$$

c).

$$Q_1=250\mu C,Q_3=-200\mu C,d_{13}=8m\\ \implies F_{13}=k\frac{250\times 200\times 10^{-12}}{64}=7.81\times 10^{-10}k=7.03N$$

d).

$$Q_2=-150\mu C,Q_3=-200\mu C,d_{23}=m\\ \implies F_{23}=k\frac{150\times 200\times 10^{-12}}{9}=30.0N$$

e).

From the free body diagram,

$$F^{Q_1}_{net}=F_{12}+F_{13}=20.53N$$

acting along the direction away from $Q_1$ to $Q_2$

f).

$$F^{Q_2}_{net}=F_{12}+F_{23}=43.5N$$

along the direction left of $Q_2$

g).

$$F^{Q_3}_{net}=F_{23}-F_{13}=22.97N$$

acting along the direction right of $Q_3$