Advertisements
Advertisements
प्रश्न
A parallel-plate capacitor has plate area 100 cm2 and plate separation 1⋅0 cm. A glass plate (dielectric constant 6⋅0) of thickness 6⋅0 mm and an ebonite plate (dielectric constant 4⋅0) are inserted one over the other to fill the space between the plates of the capacitor. Find the new capacitance.
Advertisements
उत्तर
The given system of the capacitor will behave as two capacitors connected in series.
Let the capacitances be C1 and C2.
Now,
`C_1 = (∈_0Ak_1)/d_1` and `C_2 = (∈_0Ak_2)/d_2`
Thus, the net capacitance is given by
`C = (C_1C_2)/(C_1+C_2)`
= `((∈_0Ak_1)/d_1 xx (∈_0Ak_2)/d_2)/((∈_0Ak_1)/d_1 +(∈_0Ak_2)/d_2)`
= `(∈_0A(k_1+k_2))/(k_1d_2+k_2d_1)`
= `((8.85 xx 10^-12) xx (10^-2) xx 24) / ((6 xx 4 xx 10^-3 + 4 xx 6 xx 10^-3)) = 4.425 xx 10^-11 C`
= 44.25 pF
APPEARS IN
संबंधित प्रश्न
A capacitor of capacitance ‘C’ is charged to ‘V’ volts by a battery. After some time the battery is disconnected and the distance between the plates is doubled. Now a slab of dielectric constant, 1 < k < 2, is introduced to fill the space between the plates. How will the following be affected? (a) The electric field between the plates of the capacitor Justify your answer by writing the necessary expressions.
A capacitor of capacitance ‘C’ is being charged by connecting it across a dc source along with an ammeter. Will the ammeter show a momentary deflection during the process of charging? If so, how would you explain this momentary deflection and the resulting continuity of current in the circuit? Write the expression for the current inside the capacitor.
A capacitor of capacitance C is charged to a potential V. The flux of the electric field through a closed surface enclosing the capacitor is
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
Following operations can be performed on a capacitor:
X − connect the capacitor to a battery of emf ε.
Y − disconnect the battery.
Z − reconnect the battery with polarity reversed.
W − insert a dielectric slab in the capacitor.
(a) In XYZ (perform X, then Y, then Z) the stored electric energy remains unchanged and no thermal energy is developed.
(b) The charge appearing on the capacitor is greater after the action XWY than after the action XYZ.
(c) The electric energy stored in the capacitor is greater after the action WXY than after the action XYW.
(d) The electric field in the capacitor after the action XW is the same as that after WX.
When 1⋅0 × 1012 electrons are transferred from one conductor to another, a potential difference of 10 V appears between the conductors. Calculate the capacitance of the two-conductor system.
Take `C_1 = 4.0 "uF" and C_2 = 6.0 "uF"` in figure . Calculate the equivalent capacitance of the combination between the points indicated.
Find the charge supplied by the battery in the arrangement shown in figure.
The two square faces of a rectangular dielectric slab (dielectric constant 4⋅0) of dimensions 20 cm × 20 cm × 1⋅0 mm are metal-coated. Find the capacitance between the coated surfaces.
A sphercial capacitor is made of two conducting spherical shells of radii a and b. The space between the shells is filled with a dielectric of dielectric constant K up to a radius c as shown in figure . Calculate the capacitance.
Suppose the space between the two inner shells is filled with a dielectric of dielectric constant K. Find the capacitance of the system between A and B.
An air-filled parallel-plate capacitor is to be constructed which can store 12 µC of charge when operated at 1200 V. What can be the minimum plate area of the capacitor? The dielectric strength of air is `3 xx 10^6 "Vm"^-1`
Five capacitor each of capacitance value C are connected as shown in the figure. The ratio of capacitance between P to R, and the capacitance between P and Q is ______.
Two spherical conductors A and B of radii a and b(b > a) are placed concentrically in the air. B is given a charge +Q and A is earthed. The equivalent capacitance of the system is ______.
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:
A capacitor with capacitance 5µF is charged to 5 µC. If the plates are pulled apart to reduce the capacitance to 2 µF, how much work is done?
Obtain the equivalent capacitance of the network shown in the figure. For a 300 V supply, determine the charge on each capacitor.
If the plates of a parallel plate capacitor connected to a battery are moved close to each other, then:
- the charge stored in it. increases.
- the energy stored in it, decreases.
- its capacitance increases.
- the ratio of charge to its potential remains the same.
- the product of charge and voltage increases.
Choose the most appropriate answer from the options given below:
