Physics

Capacitors and Capacitance

189 Questions

Capacitors and capacitance are crucial topics in physics, covering the storage of electric charge and energy. These concepts explore series and parallel combinations, dielectric materials, and capacitive reactance. Questions frequently appear in various competitive engineering and medical entrance exams.

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Capacitors and Capacitance Questions

Multiple choice parallel plate capacitor electrostatic potential and capacitance electrostatics physics

Choose the correct answer from the alternatives given.
The charge on a parallel plate capacitor varies as $q \, = \, q _0 \, cos2\pi \nu t$. The plates are very large and close together (area = A, separation = d). The displacement current through the capacitor is then

  1. $3q _0 \, 2\pi \nu \, sin2\pi \nu t$
  2. $-q _0 \, 2\pi \nu \, sin2\pi \nu t$
  3. $4q _0 \, 2\pi \, sin2\pi \nu t$
  4. $5q _0 \, 2\pi \, sin2\pi \nu t$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given: The charge on the parallel plate capacitor varies as $q=q _0cos2\pi\nu t$


To find: The displacement current through the capacitor.

The displacement current in a capacitor is equal to the conduction current of the capacitor.
Displacement current, $I _D \, =I _C$


The conduction current in a capacitor is given by:
$I _C= \dfrac{dq}{dt}$
$\implies\, \dfrac{d}{dt} \, [q _0 \, cos \, 2\pi\nu t]$

$  \,=\, -q _0 \, 2\pi \nu\, sin2\pi \nu\,t$


Option $(B)$ is correct.

Multiple choice parallel plate capacitor electrostatic potential and capacitance electrostatics physics

Three connected conductors A, B and C have a total charge of 48$\mu V$. The ratio of their capacitance are 1 :3 : 2. The charges on thei individually.

  1. 24$
    \mu C
    $
    , 12$
    \mu C
    $
    , 12$
    \mu C
    $
  2. 8$
    \mu C
    $
    ,18$
    \mu C
    $
    ,22$
    \mu C
    $
  3. 8$
    \mu C
    $
    ,24$
    \mu C
    $
    ,16$
    \mu C
    $
  4. 16$
    \mu C
    $
    , 16$
    \mu C
    $
    ,16$
    \mu C
    $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Three conductors $A , B , C$ have total charge $Q=48 \ \mu C$

Let charge on individual capacitors be $Q _1 , Q _2 , Q _3$
$\therefore Q _1+Q _2+Q _3=Q=48 \ \mu C ... 1$
also $C _1:C _2:C _3::1:3:2 ... 2$
For three conductors are in parallel $\dfrac{Q _1}{C _1}=\dfrac{Q _2}{C _2}=\dfrac{Q _3}{C _3} ... 3$
From $2 \  and \ 3$ we get
$3Q _1=Q _2 \ and \ 2Q _1=Q _3 ... 4$
substituting $4 \  in \ 1$
$\therefore Q _1+3Q _1+2Q _1= 48 \ \mu C$
$\therefore Q _1=8 \ \mu C$ 
substituting in $4$ we get 
$Q _2=24 \ \mu C \ and \ Q _3=16 \ \mu C$
Hence the charge on individual capacitors are $Q _1=8 \ \mu C ,Q _2=16 \ \mu C , Q _3=24 \ \mu C$

Multiple choice parallel plate capacitor electrostatic potential and capacitance electrostatics physics

Two capacitor each having a capacitance $C$ and breakdown voltage $V$ are joined in series. The effective capacitance and maximum working voltage of the combination is:-

  1. $2C, 2V$
  2. $\dfrac{C}{2}, \dfrac{V}{2}$
  3. $2C, V$
  4. $\dfrac{C}{2}, 2V$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
In series arrangement at charge on each plate of each capacitor has the same magnitude the potential difference in distributed inversely in the ratio of capacitor ie,

V=V1+V2
V=2V

The equivalent capacitance C's is given by
1/Cs=1/C1+1/C2
Cs=C/2

answer is
2V and C/2
Multiple choice parallel plate capacitor electrostatic potential and capacitance electrostatics physics

A cylindrical capacitor has two co-axial cylinders of length $20\ cm$ and radii $2r$ and $r$. Inner cylinder is given a charge $10\ \mu F$. The potential difference between the two cylinders will be ?

  1. $\dfrac {0.1 \ln {2}}{4 \pi \epsilon _{0}}m\ V$
  2. $\dfrac {\ln {2}}{4 \pi \epsilon _{0}}m\ V$
  3. $\dfrac {10\ln {2}}{4 \pi \epsilon _{0}}m\ V$
  4. $\dfrac {0.01\ln {2}}{4 \pi \epsilon _{0}}m\ V$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a cylindrical capacitor, V = (Q/(2πε₀L)) * ln(b/a). Given L=0.2m, b/a=2, and assuming Q=10μC (μF is a typo), V = (10×10⁻⁶/(2πε₀×0.2)) × ln2 = (5×10⁻⁵ ln2)/(2πε₀) ≈ (0.1 ln2)/(4πε₀) in the given notation. Option A matches this value. The 'm V' notation is unusual but represents the numerical result.

Multiple choice parallel plate capacitor electrostatic potential and capacitance electrostatics physics

A condenser of capacity $ 2 \mu F$ is charged to a potential of 200V. It is now connected to an uncharged condenser of capacity $ 3 \mu F$. The common potential is :

  1. 200 V

  2. 100 V

  3. 80 V

  4. 40 V
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Potential difference is same
$V=\dfrac{q}{C}$
$\dfrac{q _1}{C _1} = \dfrac{q _2}{C _2}=V$
$\dfrac{q _1}{2}=\dfrac{q _2}{C}=V$

By conservation of charge, $q _1+q _2=400$

Solving the above two equations give:
$q _1 = 160 \mu C$
$q _2=240 \mu C$

$V=\dfrac{q _2}{C _2}=80 V$
Multiple choice parallel plate capacitor electrostatic potential and capacitance electrostatics physics

Two connected bodies having respectively capacitances ${\text{C}} _{\text{1}} \,{\text{and}}\,{\text{C}} _{\text{2}} $ are charged with a total charge Q. The potentials of the two bodies are.

  1. $
    \dfrac{{\text{Q}}}
    {{{\text{C}} _{\text{1}} + C _2 }},\dfrac{Q}
    {{C _1 + C _2 }}
    $
  2. $
    \dfrac{Q}
    {{C _1 }} + \dfrac{{\text{Q}}}
    {{C _2 }},\dfrac{Q}
    {{C _1 }} + \dfrac{{\text{Q}}}
    {{C _3 }}
    $
  3. $
    \dfrac{{C _1 C _2 }}
    {{C _1 + C _2 }},\dfrac{{C _1 C _2 }}
    {{C _1 - C _2 }}
    $
  4. $
    \dfrac{Q}
    {{C _1 }} - \dfrac{Q}
    {{C _2 }},\dfrac{Q}
    {{C _1 - C _2 }}
    $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given that
Charge on the body $=Q$
Now, the capacitance$=C _1$ and $C _2$
Again we know that
$V=\cfrac{Q}{C}$
$\therefore$ the total capacitance here $=C _1+C _2$
$V=\cfrac{Q}{C _1+C _2}$
$\therefore$ The potential for both $=(\cfrac{Q}{C _1+C _2})(\cfrac{Q}{C _1+C _2})$

Multiple choice parallel plate capacitor electrostatic potential and capacitance electrostatics physics

Two capacitors A and B of capacitance $ 6 \mu F$ and $10 \mu F$ respectively are connected in parallel and this combination is connected in series with a third capacitors C of $ 4 \mu F $. A potential difference of 100 volt is applied across the entire combination. Find the charge and potential difference across $6\ \mu F$ capacitor.

  1. $120 \mu \, C; 20 V.$
  2. $200 \mu \, C; 20 V.$
  3. $320 \mu \, C; 80 V.$
  4. $320 \mu \, C; 60 V.$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Since A and B are connected and Parallel,
Hence,
Equivalence capacitance of capacitor$ A$ and $B = 6\mu F +10\mu F = 16\mu F$ 

Now $16\mu F$ and capacitor C of $4\mu F$ are connected in series
Hence,
the equivalence capacitance $(EC)$ will be given by
$\dfrac{1}{EC}= \dfrac {1}{16} +\dfrac {1}{4}$
we get,
Equivalence capacitance $(EC) = \dfrac{16}{5} \mu F$
$Charge = EC\times P.D.$
$Charge = \dfrac{16}{5}\times 100$
$Charge(Q) = 320\ \mu F $

Now,
$Q _A=320\times \dfrac {6}{16} \mu F$
$Q _A=120 \ \mu F$
$Q _B=320\times \dfrac {10}{16} \mu F$
$Q _B=200 \ \mu F$

Now proceeding for P.D. across each capacitor,
$V _A=\dfrac{Q _A}{C _A}$

$V _A=\dfrac{120}{6} = 20 V$ ,

$V _B=\dfrac{Q _B}{C _B}$

$V _B=\dfrac{200}{10} = 20 V$ ,

$V _C=\dfrac{Q _C}{C _C}$

$V _C=\dfrac{320}{4}=80 V$
this is the required solution.

Multiple choice parallel plate capacitor electrostatic potential and capacitance electrostatics physics

The parallel plates of capacitor are charged to a potential difference of 320 volts and are then connected across a resistor. The potential difference across the capacitor decays exponentially with time. Alter 1 second the potential difference between the plates of the capacitor is 240 volts then after 2 seconds the potential difference between the plates will be -

  1. 200 V

  2. 180 V

  3. 160 V

  4. 140 V
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The decay follows V = V0 * e^(-t/RC). At t=1, 240 = 320 * e^(-1/RC), so e^(-1/RC) = 240/320 = 0.75. At t=2, V = 320 * e^(-2/RC) = 320 * (e^(-1/RC))^2 = 320 * (0.75)^2 = 320 * 0.5625 = 180 V.

Multiple choice parallel plate capacitor electrostatic potential and capacitance electrostatics physics

Two identical capacitors are connected in parallel across a potenial difference V. after they are fully charged, the positive plate of first capacitor is connected to negative plate of second and negative plate of first is connected to positive plate of other. The loss of energy will be

  1. $\dfrac { 1 }{ 2 } { CV }^{ 2 }$
  2. ${ CV }^{ 2 }$
  3. $\dfrac { 1 }{ 4 } { CV }^{ 2 }$

  4. Zero
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice parallel plate capacitor electrostatic potential and capacitance electrostatics physics

Consider two bodies A and B of same capacitance.  If charge of -10C flows from body A to body B, then

  1. the potential of body A increases.

  2. the potential of body B decreases

  3. the magnitude of change in potential in both bodies is same.

  4. All the above
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

As we know $ V = \dfrac{Q}{C}$ , where letters have their respective meanings.

So, if charge of $-10C$  (Negative Charge) flows from body A to body B, then the potential of body A increases and the potential of body B decreases. Also the magnitude of change in potential in both bodies is same as A and B are of same capacitance.
Therefore, D is correct option.

Multiple choice parallel plate capacitor electrostatic potential and capacitance electrostatics physics

A capacitor contains two square plates with side lengths $5.0$ cm. The plates are separated by $2.0$ mm. Dry air fills the space between the plates. Dry air has a dielectric constant of $1.00$ and experiences dielectric breakdown when the electric field exceeds $3.0 \times  10^4$ V/cm.
What is the magnitude of charge that can be stored on each plate before the capacitor exceeds its breakdown limit and sends a spark between the plates?

  1. $6.6 \times 10^{-8}C$
  2. $6.6 \times 10^{-5}C$
  3. $3.3 \times 10^{-7}C$
  4. $3.3 \times 10^{-84}C$
  5. $8.1 \times 10^{-2}C$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given :   $d = 2$  mm             $E = 3.0 \times 10^4$ $V/cm = 3.0 \times 10^6$  $V/m$         $l = 5.0 cm  = 0.05$ m

Area of each plate       $A = l^2 = (0.05)^2  = 25 \times 10^{-4}$  $m^2$
Capacitance of the capacitor         $C = \dfrac{A\epsilon _o }{d}$

Potential difference between the plates         $V = Ed$
$\therefore$   Charge on each plate      $Q = CV = A\epsilon _o E$
$\implies$   $Q = 25\times 10^{-4} \times 8.85 \times 10^{-12} \times 3.0 \times 10^6  = 6.6 \times 10^{-8}$  C