How does the intensity of a sound wave change if the distance from the source increases by a factor of 2?
2 answers:
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1-2) They have same surface charge density
3-4) The metallic conductor has greatest surface charge density
Explanation:
1-2)
In a conductor, the charge carriers (mainly electrons) are free to move. Therefore, as a result, they tend to move at the largest possible distance from each other, because of the repulsive force that they exert on each other.
The configuration that maximize the distance between the charge carriers for a solid sphere of metallic conductor is the one in which all the electrons are on the surface, and they are equally spaced between each other. This means that for the solid sphere of radius R, the excess charge Q will be entirely spread over the surface of the sphere.
Similarly, the excess charge Q on the hollow spherical shell (which is also made of the same conducting material) will also be spread over the surface with the charge carriers at the maximum distance from each other. Therefore, the surface charge density for both objects will be
where R is the radius of the two spheres.
3-4)
In this case, the surface charge density on the two objects is different.
In fact, on the metallic sphere (conducting) the surface charge density is (as explained in part 1):
Hoever, the second sphere is made of an insulating material. In an insulator, the charge carriers are not free to move. If the initial charge Q is spread across the all sphere (which is not hollow), this means that some of the charge will actually also be inside the sphere. So the charge deposited on the surface, Q', will be less than the total charge Q. Therefore, the surface charge density will be
And since Q' < Q, this means that , so the conducting sphere has a greatest surface charge density.
Answer:
(a) Charge of 4.25 μF capacitor is 35.46 μC.
(b) Charge of 1.13 μF capacitor is 10.05 μC.
(c) Charge of 2.85 μF capacitor is 25.36 μC.
Explanation:
Let C₁ , C₂ and C₃ are the capacitor which are connected to the battery having voltage V. According to the problem, C₁ and C₂ are connected in parallel. There equivalent capacitance is:
C₄ = C₁ + C₂
Substitute 1.13 μF for C₁ and 2.85 μF for C₂ in the above equation.
C₄ = ( 1.13 + 2.85 ) μF = 3.98 μF
Since, C₄ and C₃ are connected in series, there equivalent capacitance is:
C₅ =
Substitute 4.25 μF for C₃ and 3.98 μF for C₄ in the above equation.
C₅ =
C₅ = 2.05 μF
The charge on the equivalent capacitance is determine by the relation :
Q = C₅ V
Substitute 2.05 μF for C₅ and 17.3 volts for V in the above equation.
Q = 2.05 μF x 17.3 = 35.46 μC
Since, the capacitors C₃ and C₄ are connected in series, so the charge on these capacitors are equal to the charge on the equivalent capacitor C₅.
Charge on the capacitor, C₃ = 35.46 μC
Charge on the capacitor, C₄ = 35.46 μC
Voltage on the capacitor C₄ = =
= 8.90 volts
Since, C₁ and C₂ are connected in parallel, the voltage drop on both the capacitors are same, that is equal to 8.90 volts.
Charge on the capacitor, C₁ = C₁ V = 1.13 μF x 8.90 = 10.05 μC
Charge on the capacitor, C₂ = C₂ V = 2.85 μF x 8.90 = 25.36 μC