Question

Two 2.00 cm diameter insulating spheres have a distance of 5.80 cm between their surfaces. The spheres each have a uniform volume charge distribution with one sphere charged to +79.6 nC and the other charged to -62.3 nC .

Part A What is the charge density of the +79.6 nC sphere? ρ = C/m3 Part B What is the electric field strength at the midpoint between the two spheres?

Answer 1

Solution:

Given:

diameter of spheres (d) = 2.0 cm = 0.02 m

radius of the spheres (r) = d/2 = 1 cm = 0.01 m

  

charge on sphere-1 : Q₁ = 79.6 x 10^-9 C

charge on sphere-2 : Q₂ = - 62.3 x 10^-9 C

distance between the spheres (d) = 5.80 cm = 0.058 m
So, distance of midpoint from center of each sphere (p) = 0.029 m

Part (A) Solution:

Charge density on  +79.6 nC sphere will be:

ρ = Q / V

Where, V = Volume of sphere = (4/3) r³ = (4/3)()(0.01 m)³ = 4.19 x 10^-6 m³

Therefore: ρ = Q / V = (79.6 x 10^-9 C) / (4.19 x 10^-6 m³) = 0.019 C/m3

Part (B) Solution:
Electric field strength due to Sphere-1 at point p will be:

total distance to point p = r + p = 0.01 m + 0.029 m = 0.039 m

Electric field: E₁ = kQ₁/r² = { (9 x10⁹)( 79.6 x 10^-9 C) } / (0.039 m) = 18369.23 N/C

Electric field strength due to sphere 2 at point p will be:

total distance to point p = r + p = 0.01 m + 0.029 m = 0.039 m

Electric field: E₂ = kQ₂/r² = { (9 x10⁹)( 62.3 x 10^-9 C) } / (0.039 m) = 14376.92 N/C

total electric field strength at mid-point 'p' will be: E = E₁ + E₂ = (18369.23 N/C) + (14376.92 N/C) = 32746.15 N/C

(Here, magnitude of both the electric field will be added, Because both the electric field will be in same direction)

Therefore: E = 3.27 x 10⁴ N/C