Electrostatic Potential and Capacitance
- A parallel plate air capacitor has capacity 'C' distance of separation between plates is 'd' and potential difference 'V' is applied between the plates. Force of attraction between the plates of the parallel plate air capacitor is :
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Force of attraction between the plates, F = qE
q × σ = q q 2∊0 2A∊0 = q² = C²V² = CV² 2 
∊0A 
× d 2cd 2d 2A∊0 Here, C = ∊0A , q = CV, A = area d Correct Option: A
Force of attraction between the plates, F = qE
q × σ = q q 2∊0 2A∊0 = q² = C²V² = CV² 2 ∊0A × d 2cd 2d 2A∊0 Here, C = ∊0A , q = CV, A = area d
- 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 dielectric constant K, which can just fill the air gap of the capacitor, is now inserted in it. Which of the following is incorrect ?
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Capacitance of the capacitor, C = Q/V
After inserting the dielectric, new capacitance C1 = K.C
New potential difference V1 = V/KUi = 1 CV² = Q² (∵ Q = CV) 2 2C Uf = Q² = Q² = C²V² = ui 2f 2KC 2KC K ∆u = uf – ui = 1 CV² 1 - 1 2 K
As the capacitor is isolated, so change will remain conserved p.d. between two plates of the capacitorL = Q = V KC K Correct Option: C
Capacitance of the capacitor, C = Q/V
After inserting the dielectric, new capacitance C1 = K.C
New potential difference V1 = V/KUi = 1 CV² = Q² (∵ Q = CV) 2 2C Uf = Q² = Q² = C²V² = ui 2f 2KC 2KC K ∆u = uf – ui = 1 CV² 1 - 1 2 K
As the capacitor is isolated, so change will remain conserved p.d. between two plates of the capacitorL = Q = V KC K
- A capacitor of 2µF is charged as shown in the diagram. When the switch S is turned to position 2, the percentage of its stored energy dissipated is :
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When S and 1 are connected
The 2µF capacitor gets charged. The potential difference across its plates will be V.
The potential energy stored in 2 µF capacitor Ui = 1 CV² = 1 × 2 × V² = V² 2 2
When S and 2 are connected
The 8µF capacitor also gets charged. During this charging process current flows in the wire and some amount of energy is dissipated as heat. The energy loss is∆U = 1 C1C2 (V1 - V2)² 2 C1 + C2
Here, C1 = 2µF, C2 8 µF, V1= V, V2 = 0∴ ΔU = 1 × 2 × 8 (v - 0)² = 4 V² 2 2 + 8 5
The percentage of the energy dissipated= ΔU × 100 = 4/5 V² × 100 = 80% Ui V² Correct Option: D
When S and 1 are connected
The 2µF capacitor gets charged. The potential difference across its plates will be V.
The potential energy stored in 2 µF capacitor Ui = 1 CV² = 1 × 2 × V² = V² 2 2
When S and 2 are connected
The 8µF capacitor also gets charged. During this charging process current flows in the wire and some amount of energy is dissipated as heat. The energy loss is∆U = 1 C1C2 (V1 - V2)² 2 C1 + C2
Here, C1 = 2µF, C2 8 µF, V1= V, V2 = 0∴ ΔU = 1 × 2 × 8 (v - 0)² = 4 V² 2 2 + 8 5
The percentage of the energy dissipated= ΔU × 100 = 4/5 V² × 100 = 80% Ui V²
