Question

A point charge of -3.00
μC is located in the center of a spherical cavity of radius 6.90 cmcm inside an insulating spherical charged solid. The charge density in the solid is 7.35 ×× 10−4−4 .C/m3

Calculate the magnitude of the electric field inside the solid at a distance of 9.50 cm from the center of the cavity.

Answer 1

Given Data

q = -3 μC = - 3 X 10^-6 C

r = 6.90 cm = 6.9 X 10^-2 m

d = 7.35 X 10⁻⁴ C/m³

distance from the center of gravity R = 9.5 cm = 9.5 X 10^-2 m

Solution : -

charge shell volume V is

V =( 4 / 3 ) ( R³ - r³ )

density = charge / volume

then charge = density X volume

charge Q = ( 4 / 3 ) ( (9.5 X 10^-2 )³ - (6.9 X 10^-2)³ ) * 7.35 X 10⁻⁴

Q = (4.18) X ( 5.29 X 10^-4 ) X 7.35 X 10⁻⁴  

Q = 1.63 X 10^-6 C

the electric field at 9.49 X 10^-2 m

E₁ = KQ / R²

E₁ = 9 X 10⁹ X 1.63 X 10^-6 / ( 9.5 X 10^-2 )²

E₁ = 1.622 X 10⁶ N/C

same way electric field at q = -3 μC

E₂ = K q / R²

E₂ = 9 X 10⁹ X (3) X 10^-6 / ( 9.5 X 10^-2 )²

E₂ = 2.99 X 10⁶ N/C

then total electric field is E

E = E₁ - E₂

E = 2.99 X 10⁶ - 1.62 X 10⁶

E = 1.368 X 10⁶ N/C

the magnitude of the electric field inside the solid at a distance of 9.5 cm from the center of the cavity is E = 1.368 X 10⁶ N/C