Practice Start Test

Capacitor

JEE Physics · 62 questions · Page 4 of 7 · Click an option or "Show Solution" to reveal answer

Q31
An electron with kinetic energy K1 enters between parallel plates of a capacitor at an angle 'α\alpha' with the plates. It leaves the plates at angle 'β\beta' with kinetic energy K2. Then the ratio of kinetic energies K1 : K2 will be :
A cos⁡2βcos⁡2α{{{{\cos }^2}\beta } \over {{{\cos }^2}\alpha }}
B cos⁡βcos⁡α{{\cos \beta } \over {\cos \alpha }}
C sin⁡2βcos⁡2α{{{{\sin }^2}\beta } \over {{{\cos }^2}\alpha }}
D cos⁡βsin⁡α{{\cos \beta } \over {\sin \alpha }}
Correct Answer
Option A
Solution

∵\because

v1cos⁡α=v2cos⁡β{v_1}\cos \alpha = {v_2}\cos \beta
v1v2=cos⁡βcos⁡α{{{v_1}} \over {{v_2}}} = {{\cos \beta } \over {\cos \alpha }}

Then the ratio of kinetic energies

k1k2=12mv1212mv22=(v1v2)2=(cos⁡βcos⁡α)2{{{k_1}} \over {{k_2}}} = {{{1 \over 2}m{v_1}^2} \over {{1 \over 2}m{v_2}^2}} = {\left( {{{{v_1}} \over {{v_2}}}} \right)^2} = {\left( {{{\cos \beta } \over {\cos \alpha }}} \right)^2}

⇒\Rightarrow

k1k2=cos⁡2βcos⁡2α{{{k_1}} \over {{k_2}}} = {{{{\cos }^2}\beta } \over {{{\cos }^2}\alpha }}
Q32
A parallel plate capacitor has 1μF capacitance. One of its two plates is given +2μC charge and the other plate, +4μC charge. The potential difference developed across the capacitor is:-
A 1V
B 5V
C 2V
D 3V
Correct Answer
Option A
Solution

Charges at inner plates are 1 μ\muC and –1 μ\muC. ∴\therefore Potential difference across capacitor =

qc=1μC1μF=1×10−6C1×10−6Farad=1V{q \over c} = {{1\mu C} \over {1\mu F}} = {{1 \times {{10}^{ - 6}}C} \over {1 \times {{10}^{ - 6}}Farad}} = 1V
Q33
Let CC be the capacitance of a capacitor discharging through a resistor R.R. Suppose t1{t_1} is the time taken for the energy stored in the capacitor to reduce to half its initial value and t2{t_2} is the time taken for the charge to reduce to one-fourth its initial value. Then the ratio t1/t2{t_1}/{t_2} will be
A 11
B 12{1 \over 2}
C 14{1 \over 4}
D 22
Correct Answer
Option C
Solution

Initial energy of capacitor,

E1=q122C{E_1} = {{q_1^2} \over {2C}}

Final energy of capacitor,

E2=12E1=q124C=(q122C)2{E_2} = {1 \over 2}{E_1} = {{q_1^2} \over {4C}} = {\left( {{{{{{q_1}} \over {\sqrt 2 }}} \over {2C}}} \right)^2}

∴\therefore

t1={t_1}=

time for the charge to reduce to

12{1 \over {\sqrt 2 }}

of its initial value and

t2={t_2} =

time for the charge to reduce to

14{1 \over 4}

of its initial value We have,

q2=q1e−t/CR{q_2} = {q_1}{e^{ - t/CR}}
⇒ln⁡(q2q1)=−tCR\Rightarrow \ln \left( {{{{q_2}} \over {{q_1}}}} \right) = - {t \over {CR}}

∴\therefore

ln⁡(12)=−t1CR...(1)\ln \left( {{1 \over {\sqrt 2 }}} \right) = {{ - {t_1}} \over {CR}}...\left( 1 \right)

and

ln⁡(14)=−t2CR  ...(2)\ln \left( {{1 \over 4}} \right) = {{ - {t_2}} \over {CR}}\,\,...\left( 2 \right)

By

(1)(1)

and

(2),(2),
t1t2=ln⁡(12)ln⁡(14){{{t_1}} \over {{t_2}}} = {{\ln \left( {{1 \over {\sqrt 2 }}} \right)} \over {\ln \left( {{1 \over 4}} \right)}}
=12ln⁡(12)2ln⁡(12)=14= {1 \over 2}{{\ln \left( {{1 \over 2}} \right)} \over {2\ln \left( {{1 \over 2}} \right)}} = {1 \over 4}
Q34
A parallel plate capacitor was made with two rectangular plates, each with a length of l=3 cml=3 \mathrm{~cm} and breath of b=1 cm\mathrm{b}=1 \mathrm{~cm}. The distance between the plates is 3μ m3 \mu \mathrm{~m}. Out of the following, which are the ways to increase the capacitance by a factor of 10 ? A. l=30 cm, b=1 cm, d=1μ ml=30 \mathrm{~cm}, \mathrm{~b}=1 \mathrm{~cm}, \mathrm{~d}=1 \mu \mathrm{~m} B. l=3 cm, b=1 cm, d=30μ ml=3 \mathrm{~cm}, \mathrm{~b}=1 \mathrm{~cm}, \mathrm{~d}=30 \mu \mathrm{~m} C. l=6 cm, b=5 cm, d=3μ ml=6 \mathrm{~cm}, \mathrm{~b}=5 \mathrm{~cm}, \mathrm{~d}=3 \mu \mathrm{~m} D. l=1 cm, b=1 cm, d=10μ ml=1 \mathrm{~cm}, \mathrm{~b}=1 \mathrm{~cm}, \mathrm{~d}=10 \mu \mathrm{~m} E. l=5 cm, b=2 cm, d=1μ ml=5 \mathrm{~cm}, \mathrm{~b}=2 \mathrm{~cm}, \mathrm{~d}=1 \mu \mathrm{~m} Choose the correct answer from the options given below:
A C only
B A only
C B and D only
D C and E only
Correct Answer
Option D
Solution

The capacitance of a parallel plate capacitor is given by

C=ϵ0Ad,C = \frac{\epsilon_0 A}{d},

where:

ϵ0\epsilon_0

is the permittivity of free space,

AA

is the area of a plate, and

dd

is the separation between the plates. A change in the capacitor’s dimensions will change its capacitance by a factor of

CnewCinitial=AnewAinitial⋅dinitialdnew.\frac{C_{\text{new}}}{C_{\text{initial}}} = \frac{A_{\text{new}}}{A_{\text{initial}}} \cdot \frac{d_{\text{initial}}}{d_{\text{new}}}.

The initial dimensions are: Length:

l=3 cm=0.03 m,l = 3\,\text{cm} = 0.03\,\text{m},

Breadth:

b=1 cm=0.01 m,b = 1\,\text{cm} = 0.01\,\text{m},

Hence, the original area is

Ainitial=0.03×0.01=3×10−4 m2,A_{\text{initial}} = 0.03 \times 0.01 = 3 \times 10^{-4}\,\text{m}^2,

Plate separation:

dinitial=3 μm=3×10−6 m.d_{\text{initial}} = 3\,\mu\text{m} = 3 \times 10^{-6}\,\text{m}.

For each modification, we compare the new factor: Case C: New dimensions:

l=6 cm=0.06 m,b=5 cm=0.05 m.l = 6\,\text{cm} = 0.06\,\text{m}, \quad b = 5\,\text{cm} = 0.05\,\text{m}.

New area:

Anew=0.06×0.05=3×10−3 m2.A_{\text{new}} = 0.06 \times 0.05 = 3 \times 10^{-3}\,\text{m}^2.

Ratio of areas:

AnewAinitial=3×10−33×10−4=10.\frac{A_{\text{new}}}{A_{\text{initial}}} = \frac{3 \times 10^{-3}}{3 \times 10^{-4}} = 10.

The plate separation is unchanged:

dinitialdnew=3×10−63×10−6=1.\frac{d_{\text{initial}}}{d_{\text{new}}} = \frac{3 \times 10^{-6}}{3 \times 10^{-6}} = 1.

Overall factor:

10×1=10.10 \times 1 = 10.

Case E: New dimensions:

l=5 cm=0.05 m,b=2 cm=0.02 m.l = 5\,\text{cm} = 0.05\,\text{m}, \quad b = 2\,\text{cm} = 0.02\,\text{m}.

New area:

Anew=0.05×0.02=1×10−3 m2.A_{\text{new}} = 0.05 \times 0.02 = 1 \times 10^{-3}\,\text{m}^2.

Ratio of areas:

AnewAinitial=1×10−33×10−4≈3.33.\frac{A_{\text{new}}}{A_{\text{initial}}} = \frac{1 \times 10^{-3}}{3 \times 10^{-4}} \approx 3.33.

New plate separation:

dnew=1 μm=1×10−6 m,d_{\text{new}} = 1\,\mu\text{m} = 1 \times 10^{-6}\,\text{m},

so the ratio of separations is

dinitialdnew=3×10−61×10−6=3.\frac{d_{\text{initial}}}{d_{\text{new}}} = \frac{3 \times 10^{-6}}{1 \times 10^{-6}} = 3.

Overall factor:

3.33×3≈10.3.33 \times 3 \approx 10.

Both Case C and Case E yield a capacitance that is 10 times the original value.

Thus, the correct modifications are those in Case C and Case E.

Q35
A parallel plate capacitor with plates of area 1 m2 each, are at a separation of 0.1 m. If the electric field between the plates is 100 N/C, the magnitude of charge on each plate is : (Take ε\varepsilon 0 = 8.85 × \times 10−-12 C2N−m2{{{C^2}} \over {N - {m^2}}})
A 9.85 × \times 10–10 C
B 8.85 × \times 10–10 C
C 6.85 × \times 10–10 C
D 7.85 × 10–10 C
Correct Answer
Option B
Solution
E=σ∈0=QA ∈0E = {\sigma \over {{ \in _0}}} = {Q \over {A\,{ \in _0}}}

Q = AE

∈\in

0 Q = (1) (100) (8.85 ×\times 10−-12) Q = 8.85 ×\times 10−-10C

Q36
A parallel plate capacitor is made of two circular plates separated by a distance 55 mmmm and with a dielectric of dielectric constant 2.22.2 between them. When the electric field in the dielectric is 3×104 V/m3 \times {10^4}\,V/m the charge density of the positive plate will be close to:
A 6×10−7  C/m26 \times {10^{ - 7}}\,\,C/{m^2}
B 3×10−7  C/m23 \times {10^{ - 7}}\,\,C/{m^2}
C 3×104  C/m23 \times {10^4}\,\,C/{m^2}
D 6×104  C/m26 \times {10^4}\,\,C/{m^2}
Correct Answer
Option A
Solution

Electric field in presence of dielectric between the two plates of a parallel plate capacitor is given by,

E=σKε0E = {\sigma \over {K{\varepsilon _0}}}

Then, charge density

σ=Kε0E\sigma = K{\varepsilon _0}E
=2.2×8.85×10−12×3×104≈6×10−7  C/m2= 2.2 \times 8.85 \times {10^{ - 12}} \times 3 \times {10^4} \approx 6 \times {10^{ - 7}}\,\,C/{m^2}
Q37
A capacitor is discharging through a resistor R. Consider in time t1, the energy stored in the capacitor reduces to half of its initial value and in time t2, the charge stored reduces to one eighth of its initial value. The ratio t1/t2 will be
A 1/2
B 1/3
C 1/4
D 1/6
Correct Answer
Option D
Solution

For a discharging capacitor when energy reduces to half the charge would become

12{1 \over {\sqrt 2 }}

times the initial value.

⇒(12)1/2=e−t1/τ\Rightarrow {\left( {{1 \over 2}} \right)^{1/2}} = {e^{ - {t_1}/\tau }}

Similarly,

(12)3=e−t2/τ{\left( {{1 \over 2}} \right)^3} = {e^{ - {t_2}/\tau }}
⇒t1t2=16\Rightarrow {{{t_1}} \over {{t_2}}} = {1 \over 6}
Q38
Capacitance of an isolated conducting sphere of radius R1 becomes n times when it is enclosed by a concentric conducting sphere of radius R2 connected to earth. The ratio of their radii (R2R1)\left( {{{{R_2}} \over {{R_1}}}} \right) is :
A nn−1{n \over {n - 1}}
B 2n2n+1{{2n} \over {2n + 1}}
C n+1n{{n + 1} \over n}
D 2n+1n{{2n + 1} \over n}
Correct Answer
Option A
Solution

Initially

=C0=4πε0R1= {C_0} = 4\pi {\varepsilon _0}{R_1}

Finally

4πε0R1R2R2−R1=nC0=4πε0nR1{{4\pi {\varepsilon _0}{R_1}{R_2}} \over {{R_2} - {R_1}}} = n{C_0} = 4\pi {\varepsilon _0}n{R_1}

⇒\Rightarrow

R2R2−R1=n{{{R_2}} \over {{R_2} - {R_1}}} = n

⇒\Rightarrow

1−R1R2=1n1 - {{{R_1}} \over {{R_2}}} = {1 \over n}

⇒\Rightarrow

R1R2=n−1n{{{R_1}} \over {{R_2}}} = {{n - 1} \over n}

⇒\Rightarrow

R2R1=nn−1{{{R_2}} \over {{R_1}}} = {n \over {n - 1}}
Q39
A slab of dielectric constant K\mathrm{K} has the same cross-sectional area as the plates of a parallel plate capacitor and thickness 34 d\dfrac{3}{4} \mathrm{~d}, where d\mathrm{d} is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be : (Given C0\mathrm{C}_{0} = capacitance of capacitor with air as medium between plates.)
A 4KC03+K\dfrac{4 K C_{0}}{3+K}
B 3KC03+K\dfrac{3 K C_{0}}{3+K}
C 3+K4KC0\dfrac{3+K}{4 K C_{0}}
D K4+K\dfrac{K}{4+K}
Correct Answer
Option A
Solution
C0=ε0Ad{C_0} = {{{\varepsilon _0}A} \over d}
C=ε0Ad−3d4+3d4K=4ε0AK3d+KdC = {{{\varepsilon _0}A} \over {d - {{3d} \over 4} + {{3d} \over {4K}}}} = {{4{\varepsilon _0}AK} \over {3d + Kd}}
=4KC03+K= {{4K{C_0}} \over {3 + K}}
Q40
A capacitor C is fully charged with voltage V0. After disconnecting the voltage source, it is connected in parallel with another uncharged capacitor of capacitance C2{C \over 2}. The energy loss in the process after the charge is distributed between the two capacitors is :
A 12CV02{1 \over 2}CV_0^2
B 14CV02{1 \over 4}CV_0^2
C 13CV02{1 \over 3}CV_0^2
D 16CV02{1 \over 6}CV_0^2
Correct Answer
Option D
Solution

Heat loss =

12(C1C2C1+C2)V02{1 \over 2}\left( {{{{C_1}{C_2}} \over {{C_1} + {C_2}}}} \right)V_0^2

=

12(C×C2C+C2)V02{1 \over 2}\left( {{{C \times {C \over 2}} \over {C + {C \over 2}}}} \right)V_0^2

=

16CV02{1 \over 6}CV_0^2
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