Electrostatic Potential And Capacitance
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A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of resulting system
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Increases by a factor of 4
-
Decreases by a factor of 2
-
Increases by a factor of 2
-
Remains the same
B.
Decreases by a factor of 2

Charge on capacitor
q = CV
when it is connected with another uncharged capacitor.

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A capacitor of 2
is charged as shown in the figure. When the switch s is turned to position 2, the percentage of its stored energy dissipated is,

-
20%
-
75%
-
80%
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0%
C.
80%
Consider the figure given above.
When switch S is connected to point 1, then initial energy stored in the capacitor is given as,
When the switch S is connected to point 2, energy dissipated on connection across 8
will be,
Therefore, per centage loss of energy = 
A parallel plate air capacitor is charged to a potential difference of V volts. After disconnecting thecharging battery the distance between the plates of the capacitor is increased using an insulating handle. As a result the potential difference between the plates
-
decreases
-
does not change
-
becomes zero
-
increases
D.
increases
Charge remains constant after charging.
If the battery is removed after charging then the charge stored in the capacitor remains constant.
q = constant
Change in capacitance

As 
Hence, 
Hence, potential difference between the plates

or 
As capacitance decreases, so potential difference increases.
A parallel plate air capacitor of capacitance C is connected to a cell of emf V and then disconnected from it. A dielectric slab of a dielectric slab of dielectric constant K, which can just fill the air gap of the capacitor, is now inserted in it. which of the following is incorrect?
-
The potential difference between the plates decreases K times
-
The energy stored in the capacitor decreases K times
-
The change in energy stored is

-
The charge on the capacitor is not conserved
C.
The change in energy stored is 

As the battery is disconnected from the capacitor the charge will not be destroyed i.e. q' = q with the introduction of dielectric in the gap of capacitor the new capacitance will be
C' = CK

A parallel plate capacitor has a uniform electrinc field E in the space between the plates. If the distance between the plates is d and area of each plate is A, the energy storedf in the capacitor is df and area of each plate is A, the energy stored in the capacitor is
C.
Energy density for a parallel plate capacitor
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Mock Test Series
Mock Test Series



