An electric field of magnitude 4.00 × 10 2 N/C exists at all points just outside the surface of a 2.00-cm-diameter steel ball bearing. Assuming the ball bearing is in electrostatic equilibrium, (a) what is the total charge on the ball? (b) What is the surface charge density on the ball?

a) 1.1115*10^-12 C, b) 3.53×10^-9 C/m²

Step-by-step explanation: according to gauss law, the electric flux from a gaussian surface is proportional to the magnitude of charge enclosed.

Mathematically, we have that

Electric flux = Q/ε0

But electric flux = EA

Hence we have that

EA = Q/ε0

Where E = strength of electric field = 4×10² N/c

A = Area of surface.

ε0 = permittivity of free space = 8.85×10^-12

Q = magnitude of charge enclosed by surface.

Let us find the area of the surface first, since diameter is 2cm, hence radius = diameter/2 = 2/2 = 1cm

Hence radius = 1cm = 0.01 m.

Since the surface is circular, area = πr²

Area (A) = 3.142×(0.01)² = 0.000314 m².

A)

Recall that

EA = Q/ε0

4×10² × 0.000314 = Q/8.85×10^-12

Q = 4×10² × 0.000314 × 8.85×10^-12

Q = 1.1115*10^-12 C

B)

Surface charge density = charge / Area

Surface charge density = 1.1115*10^-12/0.000315

Surface charge density = 3.53×10^-9 C/m²