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

Answer:

$E_{net} = 2.25 KQ\L2$

Explanation:

from figure

$d = \sqrt { l^2 -(\frac{l}{2})^2}$

  $= \sqrt {\frac{3l^2}{4}$

  $= l\sqrt{ \frac{3}{4}}$

net field at point P

$E_{net} = E_1 -E_2 +E_3$

$= K\frac{q_1}{(\frac{l}{2})^2} -K\frac{q_2}{(\frac{l}{2})^2} +K\frac{q_3}{d^2}[tex][tex]q_1 =q_2 =q_3 =27Q$

$d = l\sqrt{ \frac{3}{4}}$

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

$E = \frac{27 KQ}{\frac{3}{4}l^2}$                         {l =4l}

$E_{net} = \frac{4 *36 KQ}{3{4l}^2}$

$E_{net} = 2.25 KQ\L2$