Question

The point charges in the figure below are located at the corners of an equilateral triangle 25.0 cm on a side, where +23.0 uC and c7.45 HC. (Assume that the +x-axis is directed to the right.) (a) Find the electric field at the location of q magnitude direction N/C (counterclockwise from the axis) (b) what is the force on q', given that q" = +1.50 nC? magnitude direction 。(counterclockwise from the +x-axis)

Answer 1

Since the given triangle is equilateral, the angles are all 60 degrees

so, the net electric field in x direction at q_a is:

$E_x = k[\frac{q_b}{0.25^2}sin(60/2) - \frac{q_c}{0.25^2}sin(60/2)]$

=> $E_x = k[\frac{23\mu}{0.25^2}sin(60/2) - \frac{(-7.45\mu)}{0.25^2}sin(60/2)] = 2.1924\times 10^6 N/C$

and in y direction:

$E_y = k[\frac{q_b}{0.25^2}cos(60/2) + \frac{q_c}{0.25^2}cos(60/2)]$

=> $E_y = k[\frac{23\mu}{0.25^2}cos(60/2) - \frac{(7.45\mu)}{0.25^2}cos(60/2)] = 1.94\times 10^6 N/C$

the magnitude of the electric field will be:

$E = \sqrt{E_x^2+E_y^2} = 2.927\times 10^6 N/C$

and the direction will be:

$\theta = tan^{-1}\frac{E_y}{E_x} =42.335\degree$

b] F_x = q_aE_x = 0.003288 N

and F_y = q_aE_y = 0.00291 N

the magnitude will be:

$F = \sqrt{F_x^2+F_y^2}=4.391\times 10^{-3}N$

and the direction is:

$\theta = tan^{-1}\frac{F_y}{F_x} =42.335\degree$