Question

Work out the magnitude and direction of field at point p ,when a charge of 2uc experiences an electrical force of 5*10^2 j^ N at point p

Answer 1

hey mate
here's the solution

Answer 2

Explanation:

It is given that ,

Charge, q = 2 μC = 2 × 10⁻⁶ C

Electric force, F = 5 × 10² j N

The relation between the electric field and electric force is given by :

E = F / q

$E=\dfrac{500}{2\times 10^{-6}}$

E = 25 × 10⁷ N/C

The direction of electric field is same as the direction of electric force. Hence, the electric field is in positive y axis. i.e. j direction.

Therefore, this is the required solution.