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Which has 4 charges of equal magnitude q (some positive and some negative), at t

ID: 1536452 • Letter: W

Question


Which has 4 charges of equal magnitude q (some positive and some negative), at the comers of a square with side d. Point P is at the center of one of the sides. What is the direction of the electric field at point P? Up Down Left Right Zero What would be the direction of the electric force on a negative test charge placed at P? Up Down Left Right Zero What is the electric potential at P? (Assume that the potential is defined to be zero at a location far away from the charges.) k_cq/d -5.8 k_cq/d -k_cq/d -5.8 k_cq/d Zero

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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