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.

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.