Question

#10 a) Find the magnitude and direction of the force F21, the force of q2 acting on q1. b) Find the magnitude and direction of the force F31, the force of q3 acting on qi, then find the net force on qi c) Find the magnitude and direction of the force Fı2, the force of qi acting on q2. d) Find the magnitude and direction of the force F3, the force of q3 acting on q2, then find the net force on q2. e) Find the magnitude and direction of the force Fi3, the force of qi acting on q3. f) Find the magnitude and direction of the force F23, the force of q2 acting on qa, then find the net force on q3. 58E

Answer 1

10]

a] Net electric force on q1 is:

$F_{q1} = k\left(-\frac{q_1q_3}{(d_1+d_2)^2} - \frac{q_1q_2}{d_1^2} \right )$

$F_{q1} = k\left(-\frac{(6\mu)(-2\mu)}{(0.05)^2} - \frac{(6\mu)(1.5\mu)}{0.03^2} \right ) = -46.8 N$

the magnitude of this force is 46.8 N and the direction is to the left.

b]

Net electric force on q2 is:

$F_{q2} = k\left(\frac{q_1q_2}{d_1^2} - \frac{q_2q_3}{d_2^2} \right )$

$F_{q2} = k\left(\frac{(6\mu)(1.5\mu)}{(0.03)^2} - \frac{(1.5\mu)(-2\mu)}{0.02^2} \right ) = 157.5 N$

the magnitude of this force is 157.5 N and the direction is to the right.

c]

Net electric force on q3 is:

$F_{q3} = k\left(\frac{q_2q_3}{d_2^2} + \frac{q_1q_3}{(d_1+d_2)^2} \right )$

$F_{q3} = k\left(\frac{(1.5\mu)(-2\mu)}{(0.02)^2} + \frac{(6\mu)(-2\mu)}{0.05^2} \right ) = -110.7 N$

the magnitude of this force is 110.7 N and the direction is to the left.