Question

Calculate magnitude and direction of the electric field at a point P, whic is 30cm from a point charge of magnitude Q=-3.0×10^-6C

Answer 1

Explanation:

We want to calculate the magnitude and direction of the electric field intensity at a point P that is 30 cm from a point charge of magnitude q=3×10−6 C.

E=Fq, where E,F and q are the electric field intensity, the electrostatic force and the magnitude of the charge.

The force on a charge q1 by a charge q at a distance r is, F=14πϵ0q1qr2, where ϵ0 is the permittivity of free space.

⇒E=14πϵ0qr2=9×109(3×10−6(0.3)2)=3×105 N/C.

⇒ The magnitude of the electric field intensity at the point P is 3×105 N/C.

Answer 2

Explanation:

I hope my answer is correct