Question

A point charge of -3.00 μC is located in the center of a spherical cavity of radius 6.50 cm inside an insulating spherical charged solid. The charge density in the solid is 7.35 × 10−4 C/m3.

a) Calculate the magnitude of the electric field inside the solid at a distance of 9.50 cm from the center of the cavity. Express your answer with the appropriate units.

b) Find the direction of this electric field.

Answer 1

Given

Magnitude of the charge at the center q_p = -3.00 x 10^-6 C

Radius of the spherical cavity r = 6.50 x 10^-2 m

Density of charge ρ = 7.35 x 10^-4 C/m³

Distance d = 9.50 x 10^-2 m

Solution

A)

The charge of the solid enclosed by the Gaussian surface of radius D

Total charge enclosed by the Gaussian surface

Q = q_s+q_p

Q= 1.79 x 10^-6 + (-3.00 x 10^-6)

Q = -1.21 x 10^-6 C

Electric flux

φ = Q/ε_o

E= φ/surface area

E = φ/4πd²

E = Q/4πε_od²

E = kQ/d²

E = 9 x 10⁹ x 1.21 x 10^-6 / (9.5 x 10^-2)²

E = 1.21 x 10⁶ N/C

B)

Since the net charge enclosed by the Gaussian surface is negative the field will be directed towards the center of the cavity