Question

on an equilateral triangle, two vertices are given charges of equal magnitude but opposite polarity. What is the resultant field at the third point​

Answer 1

☆Concept:

》The direction of Electric field;

• Towards the negative charge.
• Away from the positive charge.

》Electric field is the region created by source charge, such that if any other charge present in its region will experience a force.

》Electric field is a vector quantity, its direction gives the nature of source charge.

☆Electric field by point charge:

$\boxed{\bf E = \dfrac{kq}{r²}}$

• q = source charge, r= distance where field is to find, k= elctrostatic force constant = 9×10⁹

☆Solution:

see the above image for figure!

• Net EF at 'A' due to B and C will be the vector sum if each.

•From figure:

•$\rm \vec{E}_A = \vec{E}_B + \vec{E}_C$

• $\rm \vec{E}_A = \sqrt{( \dfrac{Kq}{r²})²+(\dfrac{Kq}{r²})²+2(\dfrac{Kq}{r²})(\dfrac{Kq}{r²})cos120°}$

•$\rm \vec{E}_A = \sqrt{( \dfrac{Kq}{r²})²+(\dfrac{Kq}{r²})²+2(\dfrac{Kq}{r²})(\dfrac{Kq}{r²})(-½)}$

•$\rm \vec{E}_A = \sqrt{( \dfrac{Kq}{r²})²+(\dfrac{Kq}{r²})²-(\dfrac{Kq}{r²})²}$

•$\rm \vec{E}_A = \sqrt{( \dfrac{Kq}{r²})²}$

•$\bf \vec{E}_A = \dfrac{Kq}{r²}$

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