Advertisements
Advertisements
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.
Advertisements
उत्तर
Let 'q' be the charge on the charged capacitor. Energy stored in it is
`U=q^2/(2C)`
When another similar uncharged capacitor is connected, the net capacitance of the system is C' = 2C
The charge on the system is constant. So, the energy stored in the system now is
`U'=q^2/(2(C"))`
`=>U'=q^2/(2(2C))`
`=>U'=q^2/(4C)`
Thus, the required ratio is
`(U')/U=(q^2/(4C))/(q^2/(2C))`
`=>(U')/U=1/2`
APPEARS IN
संबंधित प्रश्न
A capacitor of unknown capacitance is connected across a battery of V volts. The charge stored in it is 300 μC. When potential across the capacitor is reduced by 100 V, the charge stored in it becomes 100 μC. Calculate The potential V and the unknown capacitance. What will be the charge stored in the capacitor if the voltage applied had increased by 100 V?
As `C = (1/V) Q` , can you say that the capacitance C is proportional to the charge Q?
A capacitor of capacitance 2⋅0 µF is charged to a potential difference of 12 V. It is then connected to an uncharged capacitor of capacitance 4⋅0 µF as shown in figure . Find (a) the charge on each of the two capacitors after the connection, (b) the electrostatic energy stored in each of the two capacitors and (c) the heat produced during the charge transfer from one capacitor to the other.
The variation of inductive reactance (XL) of an inductor with the frequency (f) of the ac source of 100 V and variable frequency is shown in fig.
- Calculate the self-inductance of the inductor.
- When this inductor is used in series with a capacitor of unknown value and a resistor of 10 Ω at 300 s–1, maximum power dissipation occurs in the circuit. Calculate the capacitance of the capacitor.
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?
Obtain the expression for capacitance for a parallel plate capacitor.
During a thunder storm, the movement of water molecules within the clouds creates friction, partially causing the bottom part of the clouds to become negatively charged. This implies that the bottom of the cloud and the ground act as a parallel plate capacitor. If the electric field between the cloud and ground exceeds the dielectric breakdown of the air (3 × 106 Vm–1), lightning will occur.
- If the bottom part of the cloud is 1000 m above the ground, determine the electric potential difference that exists between the cloud and ground.
- In a typical lightning phenomenon, around 25 C of electrons are transferred from cloud to ground. How much electrostatic potential energy is transferred to the ground?
- Charge on each capacitor remains same and equals to the main charge supplied by the battery.
- Potential difference and energy distribute in the reverse ratio of capacitance.
- Effective capacitance is even les than the least of teh individual capacitances.
Between the plates of parallel plate condenser there is 1 mm thick medium shoot of dielectric constant 4. It is charged at 100 volt. The electric field in volt/meter between the plates of capacitor is ______.
A capacitor has charge 50 µC. When the gap between the plate is filled with glass wool, then 120 µC charge flows through the battery to capacitor. The dielectric constant of glass wool is ______.
