Respuesta :
Answer:
Electric field will be [tex]394.6857\times 10^{-6}N/C[/tex]
Explanation:
We have given that a charge of [tex]-3\mu C=-3\times 10^{-6}C[/tex] is inside the cavity of sphere
Charge density of sphere [tex]\rho =7.35\times 10^{-4}C/m^3[/tex]
Radius of cavity = 6.90 cm = 0.069 m
We have to find the electric field at 9.20 cm = 0.092 m
So effective radius = 0.092 - 0.069 = 0.023 m
So effective volume [tex]V=\frac{4}{3}\pi R^3=\frac{4}{3}\times 3.14\times 0.023^3=0.5093\times 10^{-4}m^3[/tex]
We know that charge = volume × charge density
So charge [tex]Q=0.5093\times 10^{-4}\times 7.35\times 10^{-4}=374.33\times 10^{-6}C[/tex]
So net charge enclosed by sphere [tex]Q_{in}=(374.33-3)\times 10^{-6}=371.33\times 10^{-6}C[/tex]
From Gauss law we know that [tex]EA=\frac{Q_{in}}{\epsilon _0}[/tex]
[tex]E=\frac{Q_{in}}{A\epsilon _0}[/tex]
[tex]E=\frac{371.33\times 10^{-6}}{4\times 3.14\times 0.092^2\times 8.85\times 10^{-12}}=394.6857\times 10^{-6}N/C[/tex]
The magnitude of the electric field inside the solid at a distance of 9.20 cm from the center of the cavity is 394.6857x10⁻⁶ N/C.
Given to us
Charge inside the cavity = -3μC = -3x10⁻⁶ C
Charge density of the sphere, ρ = 7.35x 10⁻⁴ C/m³
Radius of he cavity = 6.90 cm = 0.0690 m
What is the Net charge electric field inside the solid at a distance of 9.20 cm from the center of the cavity?
We have to find the electric field at a radius of 9.20 cm, therefore, the effective radius,
Effective radius = 0.092-0.069 = 0.023 m
Effective Volume = [tex]\dfrac{4}{3}\pi r^3[/tex] = 0.5093 x10⁻⁴ m³
The charge, Q = volume x charge density = 374.33 x 10⁻⁶ C
The net charge enclosed,
[tex]Q_{in} = (374.33-3) \times 10^{-6} = 371.33 \times 10^{-6} \rm\ C[/tex]
What is the magnitude of the electric field inside the solid at a distance of 9.20 cm from the center of the cavity?
We know according to the Gauss law,
[tex]\overrightarrow E = \dfrac{Q_{in}}{\epsilon_o}\\\\\overrightarrow E = 394.6857 \times 10^{-6}\rm\ N/C[/tex]
Hence, the magnitude of the electric field inside the solid at a distance of 9.20 cm from the center of the cavity is 394.6857x10⁻⁶ N/C.
Learn more about Gauss Law:
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