Advertisements
Advertisements
प्रश्न
Consider the situation shown in the figure. The switch S is open for a long time and then closed. (a) Find the charge flown through the battery when the switch S is closed. (b) Find the work done by the battery.(c) Find the change in energy stored in the capacitors.(d) Find the heat developed in the system.
उत्तर
Since the switch is opened for a long time, the capacitor is in series.
`therefore C_(eq) = C/2`
When the switch is closed, the charge flown from the battery is given by
`q = C/2 xx ε =( Cε)/2`
(b) Work done, `W = q xx V`
`⇒ q = C_ε/2 xx ε = (Cε^2)/2`
(c) The change in the energy stored in the capacitors is given by
`ΔE = 1/2 C_(eq) xx V_2`
`= 1/2 xx C/2 xx ε^2 = 1/4 Cε^2`
(d) The heat developed in the system is given by
H = ΔE
`⇒ H = 1/4 Cε^2`
APPEARS IN
संबंधित प्रश्न
A parallel plate capacitor of capacitance C is charged to a potential V. It is then connected to another uncharged capacitor having the same capacitance. Find out the ratio of the energy stored in the combined system to that stored initially in the single capacitor.
Two identical capacitors of 12 pF each are connected in series across a battery of 50 V. How much electrostatic energy is stored in the combination? If these were connected in parallel across the same battery, how much energy will be stored in the combination now?
Also find the charge drawn from the battery in each case.
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 . The capacitance now becomes _________ .
Two metal spheres of capacitance C1 and C2 carry some charges. They are put in contact and then separated. The final charges Q1 and Q2 on them will satisfy
A parallel-plate capacitor has plate area 25⋅0 cm2 and a separation of 2⋅00 mm between the plates. The capacitor is connected to a battery of 12⋅0 V. (a) Find the charge on the capacitor. (b) The plate separation is decreased to 1⋅00 mm. Find the extra charge given by the battery to the positive plate.
A capacitor is made of a flat plate of area A and a second plate having a stair-like structure as shown in figure . The width of each stair is a and the height is b. Find the capacitance of the assembly.
Find the equivalent capacitance of the infinite ladder shown in figure between the points A and B.
Consider the situation shown in figure. The plates of the capacitor have plate area A and are clamped in the laboratory. The dielectric slab is released from rest with a length a inside the capacitor. Neglecting any effect of friction or gravity, show that the slab will execute periodic motion and find its time period.
A parallel plate capacitor stores a charge Q at a voltage V. Suppose the area of the parallel plate capacitor and the distance between the plates are each doubled then which is the quantity that will change?
Define ‘capacitance’. Give its unit.
Obtain the expression for energy stored in the parallel plate capacitor.
Explain in detail the effect of a dielectric placed in a parallel plate capacitor.
Calculate the resultant capacitances for each of the following combinations of capacitors.
Two similar conducting spheres having charge+ q and -q are placed at 'd' seperation from each other in air. The radius of each ball is r and the separation between their centre is d (d >> r). Calculate the capacitance of the two ball system ______.
A capacitor of 4 µ F is connected as shown in the circuit (Figure). The internal resistance of the battery is 0.5 Ω. The amount of charge on the capacitor plates will be ______.
Can the potential function have a maximum or minimum in free space?
For changing the capacitance of a given parallel plate capacitor, a dielectric material of dielectric constant K is used, which has the same area as the plates of the capacitor.
The thickness of the dielectric slab is `3/4`d, where 'd' is the separation between the plate of the parallel plate capacitor.
The new capacitance (C') in terms of the original capacitance (C0) is given by the following relation:
Two plates A and B of a parallel plate capacitor are arranged in such a way, that the area of each plate is S = 5 × 10-3 m 2 and distance between them is d = 8.85 mm. Plate A has a positive charge q1 = 10-10 C and Plate B has charge q2 = + 2 × 10-10 C. Then the charge induced on the plate B due to the plate A be - (....... × 10-11 )C
