Question

The diagram shows a point charge +q placed in the electric field of a charge +Q. The force experienced by the charge +q at point A is F. Calculate the force experienced by this charge when it is placed at points B,C,D and E. In each case, explain your answer.

Answer 1

Just attatched the diagram in a jpeg format.

Answer 2

1.The force,when the particle is at A and B are the same and is equal to KQq/R^2

2.The force,when the particle is at C and D are the same and is equal to \[KQq/(2r)^{2}\]
3.The force,when the particle is at E is \[KQq/(2\sqrt{2})^{2}\]

Answer 3

Is that the complete answer?

Answer 4

@B: since the distance of the charge q from charge Q is same ie R,the magnitude of the force experienced is the same.therefore Force @ B =F(downwards since Q repels q)
@C: the distance of the charge q from charge Q is twice as from A.As per the inverse square law the force is inversely proportional to the square of the distance ,F prop to 1/(r^2),as the distance becomes twice the dist @A ,the force reduces to one-fourth,(1/4),
FORCE @ C=F/4,(towards left),
@D: the distance of the charge q from charge Q is thrice as from A.As per the inverse square law the force is inversely proportional to the square of the distance ,F prop to 1/(r^2),as the distance becomes thrice the dist @A ,the force reduces to one-ninth,(1/9),
FORCE @ D=F/9,(towards right),
@E: the distance of the charge q from charge Q is square root 5 times as from A.As per the inverse square law the force is inversely proportional to the square of the distance ,F prop to 1/(r^2),as the distance becomes square root 5 the dist @A ,the force reduces to one-fifth,(1/5),
FORCE @ E=F/5,(diagonally at an angle of tan inverse(1/2))