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Question : 27
Total: 27
(a) Describe briefly the process of transferring the charge between the two plates of a parallel plate capacitor when connected to a battery. Derive an expression for the energy stored in a capacitor.
(b) A parallel plate capacitor is charged by a battery to a potential differenceV
OR
(a) Derive an expression for the electric field at any point on the equatorial line of an electric dipole.
(b) Two identical point charges,q 2 m Q Q
(b) A parallel plate capacitor is charged by a battery to a potential difference
OR
(a) Derive an expression for the electric field at any point on the equatorial line of an electric dipole.
(b) Two identical point charges,
Solution:
(a) Description of the process of transferring the charge.
Derivation of the expression of the energy stored.
(b) Calculation of the ratio of energy stored.
(a)
The electrons are transferred to the positive terminal of the battery from the metallic plate connected to the positive terminal, leaving behind positive charge on it. Similarly, the electrons move on to the second plate from negative terminal, hence it gets negatively charged. Process continuous till the potential difference between two plates equals the potential of the battery.
Let 'd w q ( q + d q )
d w = V d q
WhereV =
∴ d w =
d q
Total work done in changing up the capacitor
W = ∫ d w =
d q
∴ W =
Hence energy stored,
W =
( =
C V 2 =
Q V )
(b) Charge stored on the capacitorq = CV
When it is connected to the uncharged capacitor of same capacitance, sharing of charge takes place between the two capacitor till the potential of both the capacitor becomes
Energy stored on the combination,
U 2 =
C (
) 2 +
C (
) 2 =
Energy stored on single capacitor before connectingU 1 =
CV 2
Ratio of energy stored in the combination to that in the single capacitor
=
= 1 : 2
OR
(a) Derivation for the expression of the electric field on the equatorial line
(b) Finding the position and nature ofQ
(a)
The magnitude of the electric fields due to the two charges+ q q q
E + q =
E − q =
The components normal to the dipole axis cancel away and the components along the dipole axis add up.
Hence,
Total Electric field= − ( E + q + E − q ) cos θ
E =
(b)
System is in equilibrium therefore net force on each charge of system will be zero.
For the total force on 'Q
=
x = 2 − x
2 x = 2
x = 1 m
(Give full credit of this part, if a students writes directly1 m
For the equilibrium of charge "q Q q
Derivation of the expression of the energy stored.
(b) Calculation of the ratio of energy stored.
(a)
The electrons are transferred to the positive terminal of the battery from the metallic plate connected to the positive terminal, leaving behind positive charge on it. Similarly, the electrons move on to the second plate from negative terminal, hence it gets negatively charged. Process continuous till the potential difference between two plates equals the potential of the battery.
Let '
Where
Total work done in changing up the capacitor
Hence energy stored,
(b) Charge stored on the capacitor
When it is connected to the uncharged capacitor of same capacitance, sharing of charge takes place between the two capacitor till the potential of both the capacitor becomes
Energy stored on the combination,
Energy stored on single capacitor before connecting
Ratio of energy stored in the combination to that in the single capacitor
OR
(a) Derivation for the expression of the electric field on the equatorial line
(b) Finding the position and nature of
(a)
The magnitude of the electric fields due to the two charges
The components normal to the dipole axis cancel away and the components along the dipole axis add up.
Hence,
Total Electric field
(b)
System is in equilibrium therefore net force on each charge of system will be zero.
For the total force on '
(Give full credit of this part, if a students writes directly
For the equilibrium of charge "
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