Question

A +5.0-μC point charge is placed at the 0 cm mark of a meter stick and a -4.0-μC charge is placed at the 50cm mark. What is the net electric field at the 30cm mark?

Please explain why we do not take into account the negative sign of the 4uc charge. don't just workout the problem I get that its kq/r2 + kq/r2.

Answer 1

As a vector quantity, Electric field cannot be negative.

The direction of electric field is determined by the sign of q₁ and q₂. S the sign of q₂ need not be taken in calculating electric field.

Assuming that charges to be at free space,

E=5040 NC^-1 towards q_2.

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