Which has 4 charges of equal magnitude q (some positive and some negative), at t
ID: 1536452 • Letter: W
Question
Explanation / Answer
Q6. field due to positive charge is directed away from the charge and field due to negative charge is directed towards the charge.
hence field due to positive charge on the top left corner is directed to the right.
field magnitude=k*q/(d/2)^2=4*k*q/d^2
field due to top right corner negative charge is towards right.
field magnitude=k*q/(d/2)^2=4*k*q/d^2
distance from bottom charges to point P =sqrt(d^2+(d/2)^2)=1.118*d
angle made by the field due to postive charge at bottom right corner with left direction =arctan(d/(d/2))=63.435 degrees
field magnitude=k*q/(1.118*d)^2=0.8*k*q/d^2
field due to bottom left charge is directed at an angle 63.435 degrees below left direction.
field magnitude=0.8*k*q/d^2
components of the above two fields along upward and downward direction will cancel each other.
net field will be along left and with a magnitude=(2*0.8*k*q/d^2)*cos(63.435)=0.71554*k*q/d^2
so net field along right direction is more than net field along left direction.
so option D is correct answer.
Q7. force on an electric charge=charge*electric field
as chrge is neagative, force direction will be opposite to that of the electric field.
hence to the left.
option C is correct.
Q8. electric potential at P=(k*q/(d/2))+(-k*q/(d/2))+(k*q/(1.118*d))+(-k*q/(1.118*d))
=0
hence option E is correct.
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