# 2 pt charges +q and -2q are placed at the vertices of a eq. triangle ABC .obtain expression for the magnitude and direction of the resultant electric field at the vertex A due to these charges

Samyak Jain
333 Points
4 years ago
You cannot perform scalar addition of two electric field vectors. You have to do vector sum of the vectors and
then find its magnitude.
Electric field at A due to charge q at B is along side AB and away from it and that at A due to carge – 2q ta C
is along side AC and towards it.
Their magnitudes are Kq / a2 and 2Kq / a2 assuming q to be a positive charge.
Electric field due to point charge q can be resolved as (Kq / a2)cos30$\dpi{80} \degree$ j + (Kq / a2)sin30$\dpi{80} \degree$ i
= (Kq / 2a2i + ($\dpi{80} \sqrt{3}$Kq / 2a2j.
Electric field due to point charge – 2q can be resolved as – (2Kq / a2)cos30$\dpi{80} \degree$ j + (2Kq / a2)sin30$\dpi{80} \degree$ i
= (Kq / a2i  –  ($\dpi{80} \sqrt{3}$Kq / a2j.
Their resultant is (Kq / 2a2i + ($\dpi{80} \sqrt{3}$Kq / 2a2j + (Kq / a2i  –  ($\dpi{80} \sqrt{3}$Kq / a2j
=  (3Kq / 2a2i  –  ($\dpi{80} \sqrt{3}$Kq / 2a2j  =  ($\dpi{80} \sqrt{3}$Kq / 2a2)($\dpi{80} \sqrt{3}$ i – j)
Magnitude of resultant  =  ($\dpi{80} \sqrt{3}$Kq / 2a2)$\dpi{80} \sqrt{(\sqrt{3})^2 + (-1)^2}$   =   ($\dpi{80} \sqrt{3}$Kq / 2a2) . 2
=  $\dpi{80} \sqrt{3}$Kq / a2 .
Khimraj
3007 Points
4 years ago
Steps and Understanding :
1) We will find Electric Potential by writing V = K(q) / d
where q is charge contained in ring.
d - distance of charge from point .
k = 1/4π£°
where £° is permitivity in vacuum.
2) We will find Electric Field by
E = -dV/dx