Answer :
Question:
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 9.50cm from the center of the cavity.
Express your answer using two significant figures.
Answer:
The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity [tex]3.65\times 10^5N/C[/tex]
Explanation:
A point charge ,q = [tex]-2.14\times 10^{-6} C[/tex] is located in the center of a spherical cavity of radius , [tex]r =6.55\times 10^{-2}[/tex] m inside an insulating spherical charged solid.
The charge density in the solid , d = [tex]7.35 \times 10^{-4}C/m^3.[/tex]
Distance from the center of the cavity,R =[tex]9.5\times 10^{-2 }m[/tex]
Volume of shell of charge= V =[tex](\frac{4\pi}{3})[ R^3 - r^3 ][/tex]
Charge on the shell ,Q = [tex]V \times d'[/tex]
[tex]Q =(\frac{4\pi}{3})[ R^3 - r^3 ] \times d[/tex]
[tex]Q = 4.1888*\times 10^{-4 }[8.57375 - 2.81011 ]\times 7.35\times 10^{-4}[/tex]
[tex]Q = 4.1888\times 10^{-4} [5.76364 ] \times 7.35 \times 10^{-4}[/tex]
[tex]Q =2.4143 \times 10^{-4} \times 7.35 \times 10^ { -4}[/tex]
[tex]Q =1.7745 \times 10^{-6 }C[/tex]
Electric field at [tex]9.5\times 10^{-2}[/tex]m due to shell[tex]E1 = \frac{k Q}{R^2}[/tex]
E1 = [tex]\frac{ 9 \times 10^9\times 1.7745\times 10^{-6 }}{ 90.25\times 10^{-4}}[/tex]
[tex]E1 =1.769\times 10^6 N/C[/tex]
Electric field at [tex]9.5\times 10^{-2}[/tex] due to 'q' at center [tex]E2 = \frac{kq}{R^2}[/tex]
E2 =[tex]\frac{ - 9 \times 10^9\times 2.14\times 10^{-6 }}{ 90.25\times 10^{-4}}[/tex]
[tex]E2 =2.134\times 10^6 N/C[/tex]
The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity
= E2- E1
[tex]=[ 2.134 - 1.769 ]\times 10^6[/tex]
[tex]= 3.65\times 10^5 N/C[/tex]