Electrostatic Potential and Capacitance
- Two condensers, one of capacity C and other of capacity C/2 are connected to a V-volt battery, as shown.[2007] The work done in charging fully both the condensers is
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Work done = Change in energy
= 1 C + C V² = 1 3C V² = 3CV² 2 2 2 2 4 Correct Option: B
Work done = Change in energy
= 1 C + C V² = 1 3C V² = 3CV² 2 2 2 2 4
- Two parallel metal plates having charges + Q and –Q face each other at a certain distance between them. If the plates are now dipped in kerosene oil tank, the electric field between the plates will
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Electric field
E = σ = Q ε Aε
ε of kerosine oil is more than that of air.
As ε increases, E decreases.Correct Option: D
Electric field
E = σ = Q ε Aε
ε of kerosine oil is more than that of air.
As ε increases, E decreases.
- A series combination of n1 capacitors, each of value C1, is charged by a source of potential difference 4 V. When another parallel combination of n2 capacitors, each of value C2, is charged by a source of potential difference V, it has the same (total) energy stored in it, as the first combination has. The value of C2 , in terms of C1, is then
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In series, Ceff = C1 n1 ∴ Energy stored, Es = 1 Ceff C2² = 1 C1 16V² 2 2 n1 = 8V² C1 n1
In parallel, Ceff = n2 C2 Energy stored, Ep = 2 n2C2V² 2 ∴ 8V²C1 = 1 = n2C2V² n1 2 ⇒ C2 = 16C1 n1n2 Correct Option: D
In series, Ceff = C1 n1 ∴ Energy stored, Es = 1 Ceff C2² = 1 C1 16V² 2 2 n1 = 8V² C1 n1
In parallel, Ceff = n2 C2 Energy stored, Ep = 2 n2C2V² 2 ∴ 8V²C1 = 1 = n2C2V² n1 2 ⇒ C2 = 16C1 n1n2
- A parallel plate capacitor has a uniform electric 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 stored in the capacitor is :
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The energy stored by a capacitor ...(i)
U = 1 CV²...........(i) 2
V is the p.d. between two plates of the capacitor. The capacitance of the parallel plate capacitor V = E.d. C = Aε0 d
Substituting the value of C in equation (i) U = 1 Aε0 (Ed)² = 1 Aε0E²d 2 d 2 Correct Option: C
The energy stored by a capacitor ...(i)
U = 1 CV²...........(i) 2
V is the p.d. between two plates of the capacitor. The capacitance of the parallel plate capacitor V = E.d. C = Aε0 d
Substituting the value of C in equation (i) U = 1 Aε0 (Ed)² = 1 Aε0E²d 2 d 2
- Two thin dielectric slabs of dielectric constants K1 and K2 (K1 ≤ K2) are inserted between plates of a parallel plate capacitor, as shown in the figure. The variation of electric field ‘E’ between the plates with distance ‘d’ as measured from plate P is correctly shown by :
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Electric field, E ∝ 1/K
As K1 < K2 so E1 > E2
Hence graph (c) correctly dipicts the variation of electric field E with distance d.Correct Option: C
Electric field, E ∝ 1/K
As K1 < K2 so E1 > E2
Hence graph (c) correctly dipicts the variation of electric field E with distance d.