Question

Find the electric field intensity at a point 50cm

away from a charge of 5mc placed in air​

Answer 1

Answer:

50 NC–¹

Explanation:

using formula,E = 1/4π. q/r²

here,q=5 ×10-⁶C

r=30m

so, E= 9×10⁹ × 5×10-⁶/(30)²

= 5×10³×10-²

=5×10=50 NC-¹