Advertisements
Advertisements
प्रश्न
A slab of material of dielectric constant K has the same area as the plates of a parallel plate capacitor but has a thickness \[\frac{3d}{4}\]. Find the ratio of the capacitance with dielectric inside it to its capacitance without the dielectric.
Advertisements
उत्तर
\[\text { Capacitance of a capacitor without dielectric is given by }: \]
\[ C_o = \frac{\epsilon_o A}{d} . . . . . (i)\]
\[\text { Capacitance of capacitor when its plates are partly filled with dielectric of thickness t and of same area as the plates is given by }: \]
\[C = \frac{\epsilon_o A}{d - t (1 - \frac{1}{K})}\]
\[\text { Here} , t = \frac{3d}{4}\]
\[C = \frac{\epsilon_o A}{d - \frac{3d}{4}(1 - \frac{1}{K})} = \frac{\epsilon_o A}{\frac{dK + 3d}{4K}} = \frac{\epsilon_o A(4K)}{dK + 3d}\]
\[ = \frac{\epsilon_o A(4K)}{d(K + 3)} = \frac{4K}{(K + 3)} \times \frac{\epsilon_o A}{d}\]
\[\text { Therefore, the ratio of the capacitance with dielectric inside it to its capacitance without the dielectric is }\]
\[\frac{C_o}{C} = \frac{\frac{4K}{(K + 3)} \times \frac{\epsilon_o A}{d}}{\frac{\epsilon_o A}{d}} = \frac{4K}{(K + 3)}\]
APPEARS IN
संबंधित प्रश्न
Considering the case of a parallel plate capacitor being charged, show how one is required to generalize Ampere's circuital law to include the term due to displacement current.
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?
Define the capacitance of a capacitor. Obtain the expression for the capacitance of a parallel plate capacitor in vacuum in terms of plate area A and separation d between the plates.
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 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?
For a one dimensional electric field, the correct relation of E and potential V is _________.
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.
A parallel plate capacitor filled with a medium of dielectric constant 10, is connected across a battery and is charged. The dielectric slab is replaced by another slab of dielectric constant 15. Then the energy of capacitor will ______.
