The charges and coordinates of two charged particles held fixed in an xy plane a

Question

The charges and coordinates of two charged particles held fixed in an xy plane are q_1 = +3.5 mu C, x_1 =3.5 cm, y_1 = 0.50 cm and q_2 = -3.5 mu C, x_2 = -2.0 cm, y_2 = 1.5 cm. Find the magnitude and direction of the electrostatic force on particle 2 due to particle 1. Magnitude N direction degree counterclockwise from the +x-axis At what x and y coordinates should a third particle of charge q_3 = +3.5 mu C be placed such that the net electrostatic force on particle 2 due to particles 1 and 3 is zero? x-component cm y-component cm

Explanation / Answer

a)

Electric field due to particle 1 on 2 :

E = k*q1*q2/r^2

= 9*10^9*(3.5*10^-6)*(3.5*10^-6)/(((3.5 - (-2))^2 + (0.5 - 1.5)^2)*10^-4)

= 35.28 N <--------- magnitude of the force

direction = -atan((1.5-0.5)/(-2 - 3.5))

= -10.3 deg

= 169.7 deg counterclockwise from +x axis

b)

As the charges are of same magnitude, by symmetry , we can deduce that the the third particle must be placed such that the particle 2 will lie exactly between 1 and 3

So, x-coordinate of particle 3 = -2 - (3.5 + 2)= -7.5 cm

y- coordinate = 1.5 + (1.5- 0.5) = 2.5 cm

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