Advertisements
Advertisements
प्रश्न
A parallel plate capacitor (A) of capacitance C is charged by a battery to voltage V. The battery is disconnected and an uncharged capacitor (B) of capacitance 2C is connected across A. Find the ratio of total electrostatic energy stored in A and B finally and that stored in A initially.
Advertisements
उत्तर
The electric field in the region between the plates depends on the change given to the conducting plates,
For capacitor 'A' of capacitance C, having voltage V, the charge on it is given by Q = CV.
On connecting this capacitor A with capacitor B changes get distributed.
The total electrostatic energy stored in A is given by
`U_A = 1/2CV^2 = 1/2 Q^2/C` .........(i)
Since capacitor B is connected across capacitor A.
∴ `C_"net" = C_A + C_B`
= C + 2C
= 3C
∴ Total electrostatic energy in A and B is given by
`U_"net" = 1/2 Q^2/C_"net"`
= `1/2(Q^2/(3C))` ....(since total change Q remains constant)
= `1/3(1/2 Q^2/C)`
= `1/3U_A` .....[By equation (i)]
`U_"net"/U_A = 1/3`
APPEARS IN
संबंधित प्रश्न
Find the charge on the capacitor as shown in the circuit.
The energy density in the electric field created by a point charge falls off with the distance from the point charge as
(a) Find the current in the 20 Ω resistor shown in the figure. (b) If a capacitor of capacitance 4 μF is joined between the points A and B, what would be the electrostatic energy stored in it in steady state?
Two capacitors of capacitances 4⋅0 µF and 6⋅0 µF are connected in series with a battery of 20 V. Find the energy supplied by the battery.
A capacitor with stored energy 4⋅0 J is connected with an identical capacitor with no electric field in between. Find the total energy stored in the two capacitors.
A capacitor of capacitance C is given a charge Q. At t = 0, it is connected to an uncharged capacitor of equal capacitance through a resistance R. Find the charge on the second capacitor as a function of time.
A capacitor is charged by a battery and energy stored is 'U'. Now the battery is removed and the distance between plates is increased to four times. The energy stored becomes ______.
A parallel plate capacitor has a uniform electric field ‘`vec "E"`’ in the space between the plates. If the distance between the plates is ‘d’ and the area of each plate is ‘A’, the energy stored in the capacitor is ______
(ε0 = permittivity of free space)
Do free electrons travel to region of higher potential or lower potential?
In a capacitor of capacitance 20 µF, the distance between the plates is 2 mm. If a dielectric slab of width 1 mm and dielectric constant 2 is inserted between the plates, what is the new capacitance?
