Question

Point charges of 27 Q are placed at each corner of an equilateral triangle, which has sides of length 3 L. What is the magnitude of the electric field at the mid-point of any of the three sides of the triangle in units of kQ/L2?

Answer 1

`E_(net) = 2.25 KQ\L2`

Step-by-step explanation:

from figure

`d = \sqrt { l^2 -((l)/(2))^2}`

`= \sqrt {(3l^2)/(4)`

`= l\sqrt{ (3)/(4)}`

net field at point P

`E_(net) = E_1 -E_2 +E_3`

`= K(q_1)/(((l)/(2))^2) -K(q_2)/(((l)/(2))^2) +K(q_3)/(d^2)[tex]</p><p>[tex]q_1 =q_2 =q_3 =27Q`

`d = l\sqrt{ (3)/(4)}`

`E_(net)  = K(27Q)/(((l)/(2))^2) -K(27Q)/(((l)/(2))^2) +K\frac{27Q}{(l\sqrt{ (3)/(4)})^2}`

`E = (27 KQ)/((3)/(4)l^2)` {l =4l}

`E_(net) = \frac{4 *36 KQ}{3{4l}^2}`

`E_(net) = 2.25 KQ\L2`