Q#1
Suppose a charge +Q₁ is given to the positive plate and a charge -Q₂ is given to the negative plate of a capacitor. What is the "charge on the capacitor"?
Answer:
If we take the area of each facing surface = A then the total area of each plate (both faces) = 2A. The surface charge density of one plate = Q₁/2A and of the other plate = -Q₂/2A.
The directions of the electric field between the plates due to the charge on each plate is the same, hence will add up.
So, from Gauss law,
E = Q₁/(2Aε₀) + Q₂/(2Aε₀)
The potential difference between the plates, V =Ed, where d is the separation of parallel plates.
V = {Q₁/(2Aε₀) + Q₂/(2Aε₀)}d
V = (Q₁ + Q₂)d/(2Aε₀)
V = (Q₁ + Q₂)/2C, (Since C = ε₀A/d) (*)
Since the potential difference between a parallel plate capacitor having a charge Q is, V = Q/C,
Comparing it with(i) we get
Q = (Q₁ + Q₂)/2, which is the charge on the capacitor.
Q#2
As C = (1/V)Q, can you say that the capacitance is proportional to the charge Q?
Answer:
The charge on a given capacitor is proportional to the potential difference between the plates,
Q ∝ V
or, Q = CV
where C is the constant of proportionality that is called capacitance. It can be written as C = (1/V)Q but it does not mean that C is proportional to Q. Q and V both are variables and they vary such that C always remains constant for a given capacitor.
Q#3
A hollow metal sphere and a solid metal sphere of equal radii are given equal charges. Which of the two will have higher potential?
Answer:
Since the charge given to a metallic object always resides on the outer surface, both the hollow and solid spheres will have equal charges on the outer surfaces. The radius of both spheres is the same hence both of them will have equal potential.
Q#4
The plates of a parallel plate capacitor are given equal positive charges. What will be the potential difference between the plates? What will be the charge on the facing surfaces? On the outer surfaces?
Answer:
The plates are identical and the same amount and nature of charges are given to each of the plates. Hence the directions of the field due to each plate will be equal and opposite between them i.e. net field = zero.
Thus, both the plates are at the same potential and the potential difference between them is zero.
Charge on the facing surfaces:
Let us take a gaussian surface that contains a portion of a facing surface as shown in the diagram below.
E.ds = dQ/ε₀
dQ/ε₀ =0*ds = 0
dQ = 0,
So, the charge on the facing surfaces is zero.
Since the facing surfaces have zero charges, all the charge given to a plate will be on the outer surfaces.
Q#5
A Capacitor has capacitance C. Is this information sufficient to know what maximum charge the capacitor can contain? If yes, what is this charge? If no, what other information is needed?
Answer:
No, only by knowing C we can not know what maximum charge the capacitor can contain. In fact, we need to know the dielectric strength of the material between the plates. Because if a very high electric field is created between the plates, the electrons of the dielectric may get detached and it behaves like a conductor.
This is called a dielectric breakdown. The minimum electric field at which this breakdown takes place is called the dielectric strength of the dielectric. We need the value of this strength to know the maximum charge a capacitor can contain.
Q#6
The dielectric constant decreases if the temperature is increased. Explain this in terms of polarization of the material.
Answer:
With the increase in temperature the vibration of molecules of the dielectric increases. The increased vibration of the molecules reduces the alignment of molecules along the applied electric field. Thus, the polarization of dielectric decreases. So the induced electric field Eₚ also decreases. Thus, the resultant electric field inside the dielectric, E = Eₒ - Eₚ increases.
Dielectric constant, K = Eₒ/E.
So, with an increase in E, K decreases. Thus if the temperature increases dielectric constant decreases.
Q#7
When a dielectric slab is gradually inserted between the plates of an isolated parallel plate capacitor, the energy of the system decreases. What can you conclude about the force on the slab exerted by the electric field?
Answer:
Since the energy of the system decreases with the insertion of the dielectric, the decreased part of the energy must have been used in doing work by the system. As the work-done increases with the insertion of the dielectric, the direction of the force on the dielectric is sure to be along the displacement (for positive work done by the system). So, the force on the dielectric by the capacitor is attractive and parallel to the plates.
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