These solutions for Capacitors are extremely popular among Class 11 Science students for Physics Capacitors Solutions come handy for quickly completing your homework and preparing for exams. What is the "charge on the capacitor"? Also, we know that the electric field inside a capacitor is zero. Since capacitance is a proportionality constant, it depends neither on the charge on the plates nor on the potential. It only depends upon the size and shape of the capacitor and on the dielectric used between the plates. A hollow metal sphere and a solid metal sphere of equal radii are given equal charges.
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These solutions for Capacitors are extremely popular among Class 11 Science students for Physics Capacitors Solutions come handy for quickly completing your homework and preparing for exams. What is the "charge on the capacitor"? Also, we know that the electric field inside a capacitor is zero. Since capacitance is a proportionality constant, it depends neither on the charge on the plates nor on the potential.
It only depends upon the size and shape of the capacitor and on the dielectric used between the plates. A hollow metal sphere and a solid metal sphere of equal radii are given equal charges. Which of the two will have higher potential? What will be the potential difference between the plates?
What will be the charges on the facing surfaces and on the outer surfaces? It is given that the plates of the capacitor have the same charges.
In other words, they are at the same potential, so the potential difference between them is zero. Let us consider that the charge on face II is q so that the induced charge on face III is - q and the distribution is according to the figure.
Now, as point P lies inside the conductor, the total field must be zero. Thus, it seems that the whole charge given is moved to the outer surfaces, with zero charge on the facing surfaces.
A capacitor has capacitance C. Is this information sufficient to know what maximum charge the capacitor can contain? If yes, what is this charges? If no, what other information is needed? This information is not sufficient.
Since the charge is proportional to the potential difference across the capacitor, we need to know the potential difference applied across the capacitor.
The dielectric constant decreases if the temperature is increased. Explain this in terms of polarization of the material. The more aligned the molecules are with the external magnetic field, the more is the polarisation and the more will be the dielectric constant. But with increase in temperature, the thermal agitation of the molecules or the randomness in their alignment with the field increases. 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? As the energy of the system decreases, the change in the energy is negative. Force is defined as a negative rate of change of energy with respect to distance. A capacitor of capacitance C is charged to a potential V. Two capacitors each having capacitance C and breakdown voltage V are joined in series.
If the capacitors in the previous question are joined in parallel, the capacitance and the breakdown voltage of the combination will be a 2 C and 2 V b C and 2 V c 2 C and V d C and V. Therefore, the potential remains the same, that is, V. Hence, the capacitor on arm AB will not contribute to the circuit.
A dielectric slab is inserted between the plates of an isolated capacitor. The force between the plates will a increase b decrease c remain unchanged d become zero. We know that the charge is conserved in an isolated system. Thus, the force acting between the plates remains unchanged. The energy density in the electric field created by a point charge falls off with the distance from the point charge as a 1 r b 1 r 2 c 1 r 3 d 1 r 4.
A parallel-plate capacitor has plates of unequal area. The larger plate is connected to the positive terminal of the battery and the smaller plate to its negative terminal. Let Q , and Q be the charges appearing on the positive and negative plates respectively. A thin metal plate P is inserted between the plates of a parallel-plate capacitor of capacitance C in such a way that its edges touch the two plates figure Q2.
It can be observed that the charges on the plates begin to overlap each other via the metallic plate and hence begin to conduct continuously. Figure Q3 shows two capacitors connected in series and joined to a battery. The graph shows the variation in potential as one moves from left to right on the branch containing the capacitors. Now, we can see from the graph that region AB is greater than region CD. Therefore, the potential difference across capacitor C 1 is greater than that across capacitor C 2.
If the oil is pumped out, the electric field between the plates will a increase b decrease c remain the same d become zero. So, if the oil is pumped out, the electric field between the plates will increase, as the dielectric has been removed. Two metal spheres of capacitance C 1 and C 2 carry some charges. They are put in contact and then separated.
The capacitance of a capacitor does not depend on a the shape of the plates b the size of the plates c the charges on the plates d the separation between the plates. So, we can clearly see that the capacitance of a capacitor does depend on the size and shape of the plates and the separation between the plates; it does not depend on the charges on the plates. A dielectric slab is inserted between the plates of an isolated charged capacitor.
Which of the following quantities will remain the same? Thus, the net effect is a reduced electric field.
A dielectric slab is inserted between the plates of a capacitor. The charge on the capacitor is Q and the magnitude of the induced charge on each surface of the dielectric is Q '. Each plate of a parallel plate capacitor has a charge q on it. The capacitor is now connected to a batter. Now, a the facing surfaces of the capacitor have equal and opposite charges b the two plates of the capacitor have equal and opposite charges c the battery supplies equal and opposite charges to the two plates d the outer surfaces of the plates have equal charges.
C Verma the answer is a , c , d. But according to us the answer should be a , b , d all these options are the properties of a capacitor and the option c is incorrect because the battery is a source of energy not charge. Moreover if a capacitor plates have equal charge on outside and equal charge on inside then one can think that the charge on the plates must be also equal so option b cant be incorrect. The separation between the plates of a charged parallel-plate capacitor is increased.
Which of the following quantities will change? Because the charge always remains conserved in an isolated system, it will remain the same. Thus, as d increases, V increases. A parallel-plate capacitor is connected to a battery. A metal sheet of negligible thickness is placed between the plates. The sheet remains parallel to the plates of the capacitor. In other words, there will not be any change in the electric field, potential or charge.
Only, equal and opposite charges will appear on the two faces of the metal plate because of induction due to the presence of the charges on the plates of the capacitor. The amount of charge stored does not depend upon the polarity of the plates. Hence, inserting a dielectric after disconnecting the battery will not bring any change in the amount of charge stored in the capacitor. This extra work done will be dissipated as heat energy. Thus, thermal energy is developed.
However, the stored electric energy remains unchanged, that is, 1 2 CV 2. Calculate the capacitance of the two-conductor system. Calculate the charge flown through the battery. How much work has been done by the battery during the process? Find the extra charge given by the battery to the positive plate. Find the charge on each of the capacitors. How much work has been done by the battery in charging the capacitors? Now, Let the charge at each capacitor be q. Find the charge appearing on each of the three capacitors shown in figure E2.
Let us first find the equivalent capacitance. It can be observed from the circuit diagram that capacitors B and C are in parallel and are in series with capacitor A. Therefore, equal potential difference will be there on capacitor A and the system of capacitors B and C. Calculate the equivalent capacitance of the combination between the points indicated. Find the charge supplied by the battery in the arrangement shown in figure E4.
The equivalent circuit for the given case can be drawn as: It can be observed that capacitors C 1 and C 2 are in parallel. A battery of emf 10 V is connected as shown in figure E5. Find the total charge supplied by the battery to the inner cylinders. It is given that the outer cylinders are kept in contact and the inner cylinders are connected through a wire.
As the capacitors are connected in parallel, the potential difference across them is the same. Two conducting spheres of radii R 1 and R 2 are kept widely separated from each other. What are their individual capacitances? If the spheres are connected by a metal wire, what will be the capacitance of the combination?
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