Test: Energy Stored in a Capacitor (NCERT) - NEET MCQ

When battery is disconnected, charge remains constant. On introducing glass slab, capacity increases, Potential difference and energy stored decreases.

When the capacitor is connected to dc source, it gets charged. The battery is then disconnected, so no more charge can flow in. On removing dielectric capacitance decrease.
Energy stored (u = q²/2C) will increase.
Potential (V = q/C) will also increase.
Electric field (E = V/D) will increase.

Here C = 700pF = 700 × 10⁻¹²F, V = 100V
Energy stored

Here C = 16pF = 16 × 10⁻¹²FV = 80V

Here, V = 200V, q = 0.1C
Energy stored,

When the capacitor is discharged, it releases the same amount of energy i.e., 10J

Capacitance of a parallel plate capacitor is C = ε₀A/d ………(i)
Potential difference between the plates is
V = Ed
The energy stored in the capacitor is

Here r = 18cm = 18 × 10⁻²m, q = 5 × 10⁻⁶C

Energy of charged conductor is

Here, loss of energy

As the capacitors are connected in parallel, therefore potential difference across both the condensors remains the same.


Work done in charging fully both the condensors is given by

The potential energy of a charged capacitor before removing the dielectric slat is U = Q²/2C.
The potential energy of the capacitor when the dielectric slat is first removed and the reinserted in the gap between the plates is U = Q²/2C
There is no change in potential energy, therefore work done is zero.

Initial energy of the combined system

On joining the two condensers in parallel, common potential.

∴ Final energy of the combined system

Decrease in energy

Half of the energy is dissipated during charging a capacitor.

As Q = CV
∴ Initially the charge on each capacitor is
Q₁ = C₁V₁ = (3μF)(6V) = 18μC
and Q₂ = C₂V₂ = (4μF)(6V) = 24μC
When two capacitors is are joined to each other such that negative plate of one is attached with the positive plate of the other. The charges Q₁ and Q₂ are redistributed till they attain the common potential which is given by.

Common potential V = Total Charge/Total capacitance

Final energy stored,

Capacitance of a parallel plate capacitor is C = ε₀A/d ……(i)
where A is the area of each plate and d is the distance between the plates. Initial energy stored in the capacitor,

When the separation between the plates is doubled , its capcaitance becomes

As the bettary is disconnected , so charged capacitor becomes isolated and charge on it will reamain constant ,
∴ Q = Q
C'V = C'V (AsQ = CV)

Final energy stored in the capacitor ,

Required work done,