Answer:
A point charge of -2.14uC is located in the center of a spherical cavity of radius 6.55cm inside an insulating spherical charged solid. The charge density in the solid is 7.35×10−4 C/m^3.
a) Calculate the magnitude of the electric field inside the solid at a distance of 9.20cm from the center of the cavity.
Express your answer with the appropriate units.
==7.3 × 10⁻⁵N/C
Explanation:
A point charge ,q = -2.14uC = - 2.14 × 10⁻⁶
cavity of radius , r = 6.55cm = 6.55 × 10⁻²m
charge density in the solid = 7.35×10⁻⁴ C/m³
Distance from the center of the cavity,R = 9.20cm = 9.2 × 10⁻²m
Volume of shell of charge = [tex]V = \frac{4\pi }{3} (R^3 - r^3)[/tex]
Charge on the shell ,Q = V × d
[tex]Q = V = \frac{4\pi }{3} (R^3 - r^3)d[/tex]
Q = 4.1888 × ((9.2 × 10⁻²)³ - (6.55 × 10⁻²)³) × 7.35 × 10⁻⁴
Q = 1.47 × 10⁻⁶N/C
Electric field at 9.2× 10⁻²m due to shell
E1 = kQ/R²
[tex]= \frac{(9*10^9)(1.46 * 10^-^6)}{8.46 * 10^-^3 }[/tex]
= 1.563 × 10⁻⁶N/C
Electric field at 9.2 × 10⁻²m due to q at center
E2
[tex]= \frac{(9*10^9)(2.14 * 10^-^6)}{8.46 * 10^-^3 }[/tex]
= 2.276 × 10⁻⁶N/C
magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity
= E2- E1
= (2.276 × 10⁻⁶) - (1.563 × 10⁻⁶)N/C
= 0.73 × 10⁻⁶N/C
= 7.3 × 10⁻⁵N/C
