Practice Start Test

Capacitor

NEET Physics · 98 questions · Page 3 of 10 · Click an option or "Show Solution" to reveal answer

Q21
A capacitor of 2 μ\mu F is charged as shown in the diagram. When the switch S is turned to position 2, the percentage of its stored energy dissipated is
A 75%
B 80%
C 0%
D 20%
Correct Answer
Option B
Solution

Initially, the energy stored in 2 μ\muF capacitor is

Ui=12CV2=12(2×10−6)V2=V2×10−6J{U_i} = {1 \over 2}C{V^2} = {1 \over 2}\left( {2 \times {{10}^{ - 6}}} \right){V^2} = {V^2} \times {10^{ - 6}}J

Initially, the charge stored in 2 μ\muF capacitor is Q i = CV = (2 × 10 –6 )V = 2V × 10 –6 coulomb.

When switch S is turned to position 2, the charge flows and both the capacitors share charges till a common potential V C is reached.

VC=total chargetotal capacitance{V_C} = {{{\rm{total }}\,{\mathop{\rm charge}\nolimits} } \over {total\,capacitance}}
=2V×10−6(2+8)×10−6=V5volt= {{2V \times {{10}^{ - 6}}} \over {\left( {2 + 8} \right) \times {{10}^{ - 6}}}} = {V \over 5}volt

Finally, the energy stored in both the capacitors

Uf=12[(2+8)×10−6](V5){U_f} = {1 \over 2}\left[ {\left( {2 + 8} \right) \times {{10}^{ - 6}}} \right]\left( {{V \over 5}} \right)
=V25×10−6J= {{{V^2}} \over 5} \times {10^{ - 6}}J

% loss of energy,

ΔU=Ui−UfUi×100%\Delta U = {{{U_i} - {U_f}} \over {{U_i}}} \times 100\%
=(V2−V2/5)×10−6V2×10−6×100%=80%= {{\left( {{V^2} - {V^2}/5} \right) \times {{10}^{ - 6}}} \over {{V^2} \times {{10}^{ - 6}}}} \times 100\% = 80\%
Q22
A parallel plate air capacitor has capacity C, distance of separation between plates is d and potential difference VV is applied between the plates. Force of attraction between the plates of the parallel plate air capacitor is
A CV2d{{C{V^2}} \over d}
B C2V22d2{{{C^2}{V^2}} \over {2{d^2}}}
C C2V22d{{{C^2}{V^2}} \over {2d}}
D CV22d{{C{V^2}} \over {2d}}
Correct Answer
Option D
Solution

Force of attraction between the plates, F = qE

=q×σ2∈0=qq2A∈0= q \times {\sigma \over {2{ \in _0}}} = q{q \over {2A{ \in _0}}}
=q22(∈0Ad)×d=C2V22cd=CV22d= {{{q^2}} \over {2\left( {{{{ \in _0}A} \over d}} \right) \times d}} = {{{C^2}{V^2}} \over {2cd}} = {{C{V^2}} \over {2d}}

Here,

C=∈0AdC = {{{ \in _0}A} \over d}

,

q=CVq = CV

, A = area

Q23
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 dieletric constant K, which can just fill the air gap of the capacitor, is now inserted in it . Which of the following is incorrect ?
A The change in energy stored is 12CV2(1K−1){1 \over 2}C{V^2}\left( {{1 \over K} - 1} \right)
B The charge on the capacitor is not conserved.
C The potential difference between the plates decreases K times.
D The energy stored in the capaciotor decreases K times.
Correct Answer
Option B
Solution

q = CV ⇒\Rightarrow V = q/C Due to dielectric insertion, new capacitance C 2 = CK Initial energy stored in capacitor,

U1=q22C{U_1} = {{{q^2}} \over {2C}}

Final energy stored in capacitor,

U2=q22KC{U_2} = {{{q^2}} \over {2KC}}

Change in energy stored,

Δ\Delta

U = U 2 – U 1

ΔU=q22C(1K−1)=12CV2(1K−1)\Delta U = {{{q^2}} \over {2C}}\left( {{1 \over K} - 1} \right) = {1 \over 2}C{V^2}\left( {{1 \over K} - 1} \right)

New potential difference between plates

V′=qCK=VKV' = {q \over {CK}} = {V \over K}
Q24
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
A 12ε0E2{1 \over 2}{\varepsilon _0}{E^2}
B E2Adε0{{{E^2}Ad} \over {{\varepsilon _0}}}
C 12ε0E2Ad{1 \over 2}{\varepsilon _0}{E^2}Ad
D ε0EAd{\varepsilon _0}EAd
Correct Answer
Option C
Solution

Capacitance of a parallel plate capacitor is

C=ε0AdC = {{{\varepsilon _0}A} \over d}

...(i) Potential difference between the plates is V = Ed ...(ii) The energy stored in the capacitor is

U=12CV2=12(ε0Ad)(Ed)2U = {1 \over 2}C{V^2} = {1 \over 2}\left( {{{{\varepsilon _0}A} \over d}} \right){\left( {Ed} \right)^2}

(Using (i) and (ii))

=12ε0E2Ad= {1 \over 2}{\varepsilon _0}{E^2}Ad
Q25
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
A 12ε0E2{1 \over 2}{\varepsilon _0}{E^2}
B E2Adε0{{{E^2}Ad} \over {{\varepsilon _0}}}
C 12ε0E2Ad{1 \over 2}{\varepsilon _0}{E^2}Ad
D ε0EAd{\varepsilon _0}EAd
Correct Answer
Option C
Solution

Capacitance of a parallel plate capacitor is

C=ε0AdC = {{{\varepsilon _0}A} \over d}

...(i) Potential difference between the plates is V = Ed ...(ii) The energy stored in the capacitor is

U=12CV2=12(ε0Ad)(Ed)2U = {1 \over 2}C{V^2} = {1 \over 2}\left( {{{{\varepsilon _0}A} \over d}} \right){\left( {Ed} \right)^2}

(Using (i) and (ii))

=12ε0E2Ad= {1 \over 2}{\varepsilon _0}{E^2}Ad
Q26
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
A become zero
B increase
C decrease
D remain same
Correct Answer
Option C
Solution

Electric field between two parallel plates placed in vacuum is given by

E=σε0E = {\sigma \over {{\varepsilon _0}}}

In a medium of dielectric constant K,

E′=σε0E' = {\sigma \over {{\varepsilon _0}}}

For kerosene oil K > 1 ⇒\Rightarrow E' < E

Q27
A series combination of n 1 capacitors, each of value C 1 , is charged by a source of potential difference 4V. When another parallel combination of n 2 capacitors, each of value C 2 , 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 C 2 . in terms of C 1 , is then
A 2C1n1n2{{2{C_1}} \over {{n_1}{n_2}}}
B 16n2n1C116{{{n_2}} \over {{n_1}}}{C_1}
C 2n2n1C12{{{n_2}} \over {{n_1}}}{C_1}
D 16C1n1n2{{16{C_1}} \over {{n_1}{n_2}}}
Correct Answer
Option D
Solution

A series combination of n1 capacitors each of capacitance C 1 are connected to 4V source.

Total capacitance of the series combination of the capacitors is 1/C s = 1/C 1 + 1/C 1 + 1/C 1 .... upto n 1 terms = n 1 /C 1 ⇒\Rightarrow C s = C 1 /n 1 Total energy stored in a series combination of the capacitors is U s = (1/2) C s (4V) 2 = (1/2) (C 1 /n 1 ) (4V) 2 …(1) Now a parallel combination of n 2 capacitors each of capacitance C 2 are connected to V source Total capacitance of the parallel combination of capacitors is C p = C 1 + C 2 + ... + upto n 2 terms = n 2 C 2 ⇒\Rightarrow C p = n 2 C 2 Total energy stored in a parallel combination of capacitors is U p = (1/2)C p V 2 = (1/2)(n 2 C 2 )(V) 2 ...(ii) According to the given problem, U s = U p (1/2) (C 1 /n 1 )(4V) 2 = (1/2)(n 2 C 2 )(V) 2 C 1 × 16/n 1 = n 2 C 2 C 2 = 16 × C 1 / (n 1 × n 2 )

Q28
Three capacitors each of capacitance C and of breakdown voltage V are joined in series. The capacitance and breakdown voltages of the combination will be
A 3C,V33C,{V \over 3}
B C3,3V{C \over 3},3V
C 3C,3V3C,3V
D C3,V3{C \over 3},{V \over 3}
Correct Answer
Option B
Solution

In series combination of capacitors V eff = V + V + V = 3V

1Ceff=1C+1C+1C{1 \over {{C_{eff}}}} = {1 \over C} + {1 \over C} + {1 \over C}
⇒Ceff=C3\Rightarrow {C_{eff}} = {C \over 3}

Thus, the capacitance and breakdown voltage of the combination will be

C3{C \over 3}

and 3V.

Q29
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
A 12ε0E2{1 \over 2}{\varepsilon _0}{E^2}
B E2Adε0{{{E^2}Ad} \over {{\varepsilon _0}}}
C 12ε0E2Ad{1 \over 2}{\varepsilon _0}{E^2}Ad
D ε0EAd{\varepsilon _0}EAd
Correct Answer
Option C
Solution

Capacitance of a parallel plate capacitor is

C=ε0AdC = {{{\varepsilon _0}A} \over d}

...(i) Potential difference between the plates is V = Ed ...(ii) The energy stored in the capacitor is

U=12CV2=12(ε0Ad)(Ed)2U = {1 \over 2}C{V^2} = {1 \over 2}\left( {{{{\varepsilon _0}A} \over d}} \right){\left( {Ed} \right)^2}

(Using (i) and (ii))

=12ε0E2Ad= {1 \over 2}{\varepsilon _0}{E^2}Ad
Q30
Two condensers, one of capacity C and other of capacity C/2 are connected to a V-volt battery, as shown in the figure. The work done in charging fully both the condensers is
A 14CV2{1 \over 4}C{V^2}
B 34CV2{3 \over 4}C{V^2}
C 12CV2{1 \over 2}C{V^2}
D 2CV 2
Correct Answer
Option B
Solution

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

∴\therefore

Q1=CV{Q_1} = CV
Q2=C2V{Q_2} = {C \over 2}V

Aslo,

Q=Q1+Q2Q = {Q_1} + {Q_2}
=CV+Q2=32CV= CV + {Q_2} = {3 \over 2}CV

Work done in charging fully both the condensors is given by

W=12QV=12×(32CV)V=34CV2W = {1 \over 2}QV = {1 \over 2} \times \left( {{3 \over 2}CV} \right)V = {3 \over 4}C{V^2}
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