Respuesta :
Answer:
Electric field at radius r inside the solid sphere is
[tex]E=\dfrac{q_{1}r}{4\pi \epsilon_{o} r_{1}^3}\ N/C\\[/tex]
Electric field at radius r between inner radius and outer radius of the shell is
E=0 N/C
Explanation:
Given
The radius of the solid sphere is [tex]r_{i}=0.105 \ m\\[/tex]
The charge on the solid sphere is [tex]q_{1}=+4.4\ \mu C[/tex]
The inner radius of the shell is [tex]r_{3}=0.15\ m[/tex]
The outer radius of the shell is [tex]r_{3}=0.242\ m[/tex]
The total charge on the shell is [tex]q_{2}=-1.1\ \mu C[/tex]
PART(A)
The magnitude of electric field at radius r where [tex] r<r_{1}[/tex] \\The volumetric charge density of the solid sphere will be
[tex]\rho=\dfrac{q_{1}}{V}\\ \rho=\dfrac{q_{1}}{\dfrac{4}{3}\pi r_{1}^3}\\[/tex]
The charge enclosed by the radius r inside the solid sphere is
rho=[tex]q_{enc}=\rho\times \dfrac{4}{3}\pi r^3\\q_{enc}=\dfrac{q_{1}}{\dfrac{4}{3}\pi r_{1}^3}\times \dfrac{4}{3}\pi r^3\\q_{enc}=\dfrac{q_{1}r^3}{r_{1}^3}\\[/tex]
According to gauss law
[tex]EA=\dfrac{q_{enc}}{\epsilon_{o}}\\E=\dfrac{\dfrac{q_{1}r^3}{r_{1}^3}}{\epsilon_{o}\times 4\pi r^2}\\E=\dfrac{q_{1}r}{4\pi \epsilon_{o} r_{1}^3}\ N/C\\[/tex]
PART(B)
The electric field at radius r where [tex]r_{2}<r<r_{3}[/tex]
The shell is conducting so due to induction of charge
The charge induced on the inner surface of the shell is equal in magnitude of the total charge on the solid sphere but polarity will be changed because a conducting shell has no net electric field inside the shell
So,
The charge on the inner surface of the shell is
[tex]q_{i}=-q_{1}\\q_{i}=-4.4\ \mu C\\[/tex]
Due to conservation of the charge on the shell
The charge accumulated on the outer surface of the shell is
[tex]q_{o}=q_{2}-q_{i}\\q_{o}=-1.1\ \mu C-(-4.4\ \mu C)\\q_{o}=-1.1\ \mu C+4.4\ \mu C\\q_{o}=3.3\ \mu C\\[/tex]
The charge enclosed by the radius r where [tex]r_{1}<r<r_{2}[/tex]
[tex]q_{enc}=q_{1}+q_{i}\\q_{enc}=4.4\times 10^{-6}-4.4\times 10^{-6}\\ q_{enc}=0\\[/tex]
According to gauss law
[tex]EA=\dfrac{q_{enc}}{\epsilon_{o}}\\ E=0\ N/C\\[/tex]
Solid aluminum sphere which is electric charged and surrounded by a spherical copper shell has,
- (a) The magnitude of the electric field, inside the aluminum ball is 2.93*10⁸.
- (b) The magnitude of the electric field, inside the copper of the shell, such that r2 < r < r3 is 0.
What is electric field?
The electric field is the field, which is surrounded by the electric charged. The electric field is the electric force per unit charge.
Given information-
The given data in the problem is,
- The radius of the solid aluminum sphere is 0.105 m.
- The charge on this sphere is +4.4 μC.
- The inner radius of the spherical copper shell is 0.15 m.
- The outer radius of the spherical copper shell is 0.242 m.
- The charge place on copper shell is -1.1 μC.
- (a) The magnitude of the electric field, in newtons per coulomb, inside the aluminum ball-
The electric field of the solid sphere inside the sphere for (r<r₁) can be given as,
[tex]E=\dfrac{Qr}{4\pi\varepsilon_0 r_1^3}[\tex]
Put the values we get,
[tex]E=\dfrac{4.4\times0.105}{4\pi\times8.85\times10^{-12}\times 0.242^3}=2.93\times10^8[\tex]
- (b) The magnitude of the electric field, inside the copper of the shell, i.e., at a radius r from the center such that r2 < r < r3-
The magnitude of the electric field inside the copper shell is zero, as the charge outside the radius (r), does not affect the electric field at (r).
- (a) The magnitude of the electric field, inside the aluminum ball is 2.93*10⁸.
- (b) The magnitude of the electric field, inside the copper of the shell, such that r2 < r < r3 is 0.
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