Electrostatic Potential and Capacitance

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Question
CBSEENPH12040111

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

  • Increases by a factor of 4

  • Decreases by a factor of 2

  • Increases by a factor of 2

  • Remains the same

Solution

B.

Decreases by a factor of 2


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

straight V subscript straight c space equals space fraction numerator straight q subscript 1 plus straight q subscript 2 over denominator straight C subscript 1 plus straight C subscript 2 end fraction space equals space fraction numerator straight q plus 0 over denominator straight C plus straight C end fraction
straight V subscript straight c space equals space straight V over 2
Initial space energy
straight U subscript straight i space equals space 1 half CV squared
Final space energy
straight U subscript straight f space equals 1 half straight C open parentheses straight V over 2 close parentheses squared plus 1 half straight C open parentheses straight V over 2 close parentheses squared
space equals open parentheses space CV squared over 4 close parentheses
Loss space of space energy space equals straight U subscript straight i space minus straight U subscript straight f
space equals space CV squared over 4

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Question
CBSEENPH12039778

A capacitor of 2μF 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%

  • 0%

Solution

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,
straight E space equals space 1 half open parentheses 2 μF close parentheses xV squared
When the switch S is connected to point 2, energy dissipated on connection across 8μF will be,
straight E apostrophe space equals space 1 half open parentheses fraction numerator straight C subscript 1 straight C subscript 2 over denominator straight C subscript 1 plus straight C subscript 2 end fraction close parentheses. space straight V squared
space space space space space equals space 1 half straight x fraction numerator 2 μFx 8 μF over denominator 10 μF end fraction space straight x space straight V squared
space space space space space equals 1 half straight x space left parenthesis 1.6 space μF right parenthesis space straight x space straight V squared
Therefore, per centage loss of energy = fraction numerator 1.6 over denominator 2 end fraction x 100 space equals space 80 percent sign

Question
CBSEENPH12040104

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

Solution

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
                  straight C apostrophe space equals space fraction numerator straight E subscript 0 straight A over denominator straight d apostrophe end fraction
As                straight d apostrophe greater than straight d
Hence,        straight C apostrophe less than straight C
Hence, potential difference between the plates
                       straight V apostrophe space equals space fraction numerator straight q over denominator straight C apostrophe end fraction
or      DV apostrophe space proportional to space space fraction numerator 1 over denominator straight C apostrophe end fraction
As capacitance decreases, so potential difference increases.

Question
CBSEENPH12039797

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 1 half C V squared space open parentheses 1 over K minus 1 close parentheses

  • The charge on the capacitor is not conserved

Solution

C.

The change in energy stored is 1 half C V squared space open parentheses 1 over K minus 1 close parentheses

When space straight a space parallel space plate space air space capacitor space connected space to space cell space of space emf space straight V comma space
then space charge space stored space will space be
straight q space equals space CV
rightwards double arrow straight V space equals space straight q over straight C
Also space energy space stored space is space straight U space equals 1 half CV squared space equals fraction numerator straight q squared over denominator 2 straight C end fraction
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
rightwards double arrow space space space space space straight V apostrophe space equals space fraction numerator straight q over denominator straight C apostrophe end fraction space equals space straight q over straight C
The space new space energy space stored space will space be space
straight U to the power of apostrophe space equals space fraction numerator straight q squared over denominator 2 CK end fraction
increment straight U space equals space straight U to the power of apostrophe minus straight U space equals space fraction numerator straight q squared over denominator 2 straight C end fraction open parentheses 1 over straight K minus 1 close parentheses
space equals space 1 half CV squared space open parentheses 1 over straight K minus 1 close parentheses

Question
CBSEENPH12039821

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 

  • 1 half straight epsilon subscript straight o straight E squared
  • straight E squared Ad divided by straight epsilon subscript straight o
  • 1 divided by 2 space straight epsilon subscript straight o straight E squared Ad
  • straight epsilon subscript straight o EAd

Solution

C.

1 divided by 2 space straight epsilon subscript straight o straight E squared Ad

 Energy density for a parallel plate capacitor
equals 1 half straight epsilon subscript straight o straight E squared space and space volume space equals space Ad
Total space energy space equals space energy space density space straight x space volume
space equals space open parentheses 1 half straight epsilon to the power of straight o straight E squared close parentheses space straight x space left parenthesis Ad right parenthesis
space equals space 1 half straight epsilon subscript straight o straight E squared Ad

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