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Grade: 12
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
8 months ago

Answers : (2)

Samyak Jain
333 Points
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\degree j + (Kq / a2)sin30\degree i
                                                                          = (Kq / 2a2i + (\sqrt{3}Kq / 2a2j.
Electric field due to point charge – 2q can be resolved as – (2Kq / a2)cos30\degree j + (2Kq / a2)sin30\degree i
                                                                         = (Kq / a2i  –  (\sqrt{3}Kq / a2j.
Their resultant is (Kq / 2a2i + (\sqrt{3}Kq / 2a2j + (Kq / a2i  –  (\sqrt{3}Kq / a2j
                        =  (3Kq / 2a2i  –  (\sqrt{3}Kq / 2a2j  =  (\sqrt{3}Kq / 2a2)(\sqrt{3} i – j)
Magnitude of resultant  =  (\sqrt{3}Kq / 2a2)\sqrt{(\sqrt{3})^2 + (-1)^2}   =   (\sqrt{3}Kq / 2a2) . 2
                                      =  \sqrt{3}Kq / a2 .
8 months ago
3008 Points
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 
8 months ago
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