Respuesta :
Answer:
(a) We should apply Gauss' Law to each region to find the corresponding electric field. In order to use Gauss' Law, we should draw an imaginary spherical surface inside each region. The electric field passes through this imaginary surface is equal to the charge inside in this surface.
[tex]\int \vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}[/tex]
The integral in the left-hand side is equal to the area of the imaginary surface.
(i) r < a:
E = 0, since there is no charge inside the imaginary surface.
(ii) a < r < b:
E = 0, since the electric field is always zero within conductors. Another reason for this is that the inner shell has no charge, and the enclosed charge in the Gauss' Law is zero.
(iii) b < r < c:
The imaginary surface is drawn in this region and it contains +2q charge, which is the net charge of the smaller shell. Therefore, Gauss' Law gives
[tex]E4\pi r^2 = \frac{2q}{\epsilon_0}\\E = \frac{q}{2\pi \epsilon_0 r^2}[/tex]
(iv) c < r < d:
The enclosed charge inside the imaginary surface in this region is the sum of the total charge of the smaller shell and the inner surface of the larger shell, which are +2q - 2q = 0. Equivalently, the electric field within a conductor is zero always.
E = 0.
(v) r > d:
The imaginary surface is now drawn to the outermost region and contains the total charge of both spheres, which is +6q. Therefore, Gauss' Law reads
[tex]E4\pi r^2 = \frac{6q}{\epsilon_0}\\E = \frac{3q}{2\pi\epsilon_0 r^2}[/tex]
(b)
Since both shells are conductors, the charges are distributed around the inner and outer surfaces, so no charge within the conductor.
(i) Since the innermost region (r > a) has no charge, the inner surface of the smaller shell is also neutral.
(ii) Since the total charge of the smaller shell is +2q, and the inner surface is neutral, the outer surface has a charge of +2q.
(iii) The +2q charge on the outer surface of the smaller shell attracts an equal amount of opposite charges on the inner surface of the larger shell. So, the inner surface of the larger shell has a charge of -2q.
(iv) Since the total charge of the larger shell is +4q, and the inner surface contains -2q, then the outer surface has a charge of +6q.
A) The electric field ( E ) in terms of q and r
i) E = 0
ii) E = 0
iii) E = [tex]\frac{q}{2\pi e_{0}r^{2} }[/tex]
iv) E = 0
v) E = [tex]\frac{3q}{2\pi e_{0}r^{2} }[/tex]
B) The Total charge on the following
i) Inner surface of the small shell while have a total charge = 0
given that ( r > a)
ii) outer surface of the small shell will have a total charge = +2q
given that inner surface charge = 0
iii) Inner surface of the large shell has a charge of -2q
iv) Outer surface of the Large shell will have a charge of +6q given that total charge on larger shell = +4q and the inner surface charge = -2q
Given data :
Inner radius ( c ) = +2q
outer radius ( d ) = +4q
A) Determine The electric field ( E ) in terms of q
To determine the electric field we will apply Gauss' Law to each region
Gauss' Law = ∫ Eda = [tex]\frac{Q}{e_{0} }[/tex] . where ∫ Eda = area of imaginary surface
i) when r < a
The electric field ( E ) = 0. This is due to the absence of charge in the imaginary surface.
ii) when a < r < b
Electric field ( E ) = 0. This is because of the absence of charge in the inner shell also enclosed charge in Gauss Law = 0
iii) When b < r < c
The electric field ( E ) = [tex]\frac{q}{2\pi e_{0}r^{2} }[/tex]. This is because has +2q charge and the imaginary surface is drawn within the region where b< r < c.
iv) when c < r < d
The electric field ( E ) = 0. This is because The sum total of the charge within the enclosed imaginary surface in the region is +2q - 2q = 0.
v) when r > d
In this region the imaginary surface is drawn up to the outermost region and will contain a total charge of +6q
The electric field ( E ) = [tex]\frac{3q}{2\pi e_{0}r^{2} }[/tex]
B) Determine the total charge on the
note ; both shells are electrical conductors
i) Inner surface of the small shell while have a total charge = 0
given that ( r > a)
ii) outer surface of the small shell will have a total charge = +2q
given that inner surface charge = 0
iii) Inner surface of the large shell has a charge of -2q
iv) Outer surface of the Large shell will have a charge of +6q given that total charge on larger shell = +4q and the inner surface charge = -2q
Hence we can conclude that the answers to your questions are as listed above
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