Asmall glass bead has been charged to 8.0 nc. what is the magnitude of the electric field 2.0 cm from the center of the bead? (k = 1/4δή0 = 8.99 × 109 n · m2/c2)a) 3.6 × 10-6 n/c b) 1.4 × 10-3 n/c c) 1.8 × 105 n/c d) 3.6 × 103 n/c

C) 1.8×10⁵ N/C.

Explanation:

Electric field: This can be defined as the region where electric force is experienced.

Electric Field intensity: This is defined as the force per unit charge which it exert at that point.

The S.I unit of electric field is N/C.

Mathematically, Electric Field intensity can be represented as,

E = kq/r² Equation 1

Where E = electric field intensity, q = charge, r = distance. k = proportionality constant.

Given: q = 8.0 nC = 8×10⁻⁹ C, r = 2.0 cm = 0.02 m, k = 8.99×10⁹ Nm²/C²

Substituting into equation 1

E = (8.99×10⁹×8×10⁻⁹)/0.02²

E = 71.92/0.0004

E = 1.798×10⁵ N/C.

E ≈ 1.8×10⁵ N/C.

The right option is C) 1.8×10⁵ N/C.