Question

A charge of +2.5 C is placed 2.0 cm from the small object. What is the magnitude of the force on the charge? Will it be attractef or repealed by the object?

the electric field 2.0 cm from a small object points away from the object with a strength of 270,000 N/C

Answer 1

Given the electric field at a distance 2cm from the small object is

E = 270000 N/C

The electric force on charge q in presence of electric field E is

F = qE

F = (+2.5 C) × (270000 N/C)

The given charge (+2.5 C) is a positive charge. It is given that the electric field of small object points away from the object. We know that electric field of a positive charge points away from the charge. Therefore, the small object is also positively charged.

Since two positive charges repel each other, +2.5 C charge will be repelled by the small object.