Question

What is the magnitude of point charge so that electric field 20cm away from it has a magnitude of 18×10 6

Answer 1

Answer:

Explanation:

r = 50cm = 0.5m

E = 2N/C

so, 2 = 9 × 10^9Q/(0.5)²

or, 2 × 0.5 × 0.5 = 9 × 10^9 Q

or, 0.5/9 × 10^-9 = Q

or, Q = 0.055 × 10^-9 = 5.5 × 10^-11 C