Question

A small glass bead has been charged to 7.6 nC. What is the magnitude of the electric field 2.0 cm from the center of the bead? (k = 1/4πε 0 = 8.99 × 109 N · m2/C2)

Answer 1

1.7 × 10^8 N/C

Step-by-step explanation:

Q = 7.6 nC, r = 2 cm = 0.02 m

E = KQ/r^2

E =

8.99 × 10^9 × 7.6 × 10^(-6)/(0.02 × 0.02)

E = 1.7 × 10^8 N/C