Advertisements
Advertisements
प्रश्न
A parallel-plate capacitor is charged to a potential difference V by a dc source. The capacitor is then disconnected from the source. If the distance between the plates is doubled, state with reason how the following change:
(i) electric field between the plates
(ii) capacitance, and
(iii) energy stored in the capacitor
Advertisements
उत्तर
(i)
Q = CV
`Q = ((epsi_0A)/d) (Ed)`
`Q = epsi_0AE`
`therefore E =Q/(epsi_0A)`
Therefore, the electric field between the parallel plates depends only on the charge and the plate area. It does not depend on the distance between the plates.
Since the charge as well as the area of the plates does not change, the electric field between the plates also does not change.
(ii)
Let the initial capacitance be C and the final capacitance be C'.
Accordingly,
`C = (epsiA)/d`
`C' = (epsi_0A)/(2d)`
`C/C' = 2`
`C' \ C/2`
Hence, the capacitance of the capacitor gets halved when the distance between the plates is doubled.
(iii)
Energy of a capacitor, U `=1/2 (Q_2)/C`
Since Q remains the same but the capacitance decreases,
`U' = 1/2 (Q^2)/((C/2))`
`U/U' = 1/2`
U' = 2U
The energy stored in the capacitor gets doubled when the distance between the plates is doubled.
APPEARS IN
संबंधित प्रश्न
In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10−3 m2 and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor?
Show that the force on each plate of a parallel plate capacitor has a magnitude equal to `(1/2)` QE, where Q is the charge on the capacitor, and E is the magnitude of the electric field between the plates. Explain the origin of the factor `1/2`.
A ray of light falls on a transparent sphere with centre C as shown in the figure. The ray emerges from the sphere parallel to the line AB. Find the angle of refraction at A if the refractive index of the material of the sphere is \[\sqrt{3}\].
A slab of material of dielectric constant K has the same area as that of the plates of a parallel plate capacitor but has the thickness d/2, where d is the separation between the plates. Find out the expression for its capacitance when the slab is inserted between the plates of the capacitor.
A parallel-plate capacitor of plate area 40 cm2 and separation between the plates 0.10 mm, is connected to a battery of emf 2.0 V through a 16 Ω resistor. Find the electric field in the capacitor 10 ns after the connections are made.
A parallel-plate capacitor has plate area 20 cm2, plate separation 1.0 mm and a dielectric slab of dielectric constant 5.0 filling up the space between the plates. This capacitor is joined to a battery of emf 6.0 V through a 100 kΩ resistor. Find the energy of the capacitor 8.9 μs after the connections are made.
A parallel plate air condenser has a capacity of 20µF. What will be a new capacity if:
1) the distance between the two plates is doubled?
2) a marble slab of dielectric constant 8 is introduced between the two plates?
Two identical capacitors are joined in parallel, charged to a potential V, separated and then connected in series, the positive plate of one is connected to the negative of the other. Which of the following is true?
A parallel plate capacitor is connected to a battery as shown in figure. Consider two situations:
- Key K is kept closed and plates of capacitors are moved apart using insulating handle.
- Key K is opened and plates of capacitors are moved apart using insulating handle.
Choose the correct option(s).
- In A: Q remains same but C changes.
- In B: V remains same but C changes.
- In A: V remains same and hence Q changes.
- In B: Q remains same and hence V changes.
Two charges – q each are separated by distance 2d. A third charge + q is kept at mid point O. Find potential energy of + q as a function of small distance x from O due to – q charges. Sketch P.E. v/s x and convince yourself that the charge at O is in an unstable equilibrium.
