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

An air-gap parallel plate capacitor is fully charged by a battery.
What combination of two measurements will allow someone to calculate the magnitude of the electric field in between the capacitor plates?

  1. The potential difference of the battery and the area of the plates.

  2. The charge on the plates and the distance between the plates.

  3. The charge on the plates and the area of the plates.

  4. The area of the plates and the distance between the plates.

  5. More than two measurements are needed to calculate the electric field in between the capacitor plates.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Capacitance of the parallel plate capacitor       $C = \dfrac{A\epsilon _o}{d}$

Using     $Q = CV$               $\implies V = \dfrac{Q}{C}$
 Electric field between the plates      $E = \dfrac{V}{d}$
$\therefore$   $E =\dfrac{Q}{C d} = \dfrac{Q}{\frac{A\epsilon _o}{d} \times d} = \dfrac{Q}{A\epsilon _o}$
Thus option C is correct.

Multiple choice parallel plate capacitor electrostatic potential and capacitance electrostatics physics

Two capacitors of $10\ pF$ and $20\ pF$ are connected to $200\ V$ and $100\ V$ sources respectively. If they are connected in parallel by the wire, what is the common potential of the capacitors?

  1. $133.3\ Volt$
  2. $150\ Volt$
  3. $300\ Volt$
  4. $400\ Volt$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Total charge on the two capacitors is:

$Q = Q _1 +Q _2 = C _1V _1+C _2V _2 $
$Q = 10 \times 10^{-9} \times 200 + 20 \times 10^{-9} \times 100$
$Q = 4\ \mu C$

Net capacitance of two capacitors in parallel is:
$C = C _1+C _2$
$C = 30\ pF$

Common potential of the parallel combination of capacitors is:
$V = \cfrac{Q}{C}$
$V = \cfrac{4\times 10^{-6}}{30 \times 10^{-9}} = 133.3\ V$

Multiple choice parallel plate capacitor electrostatic potential and capacitance electrostatics physics

A parallel plate capacitor has an electric field of $105$V /m between the plates .If the charge on one of the capacitor plate is 1$\mu$C,then the magnitude of the force on each capacitor plate is :

    1. 1 N
    1. 05 N
    1. 5 N
    1. 01 N
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Force of attraction between the plates capacitor 

$F=\cfrac {1}{2} Q({\dfrac{Q}{A\epsilon _0} })$

=$\cfrac{1}{2} QE$

$\dfrac{1}{2}\times 10^{-6}\times 10^5$

$\dfrac{1}{20}$

$=0.05 N$

Multiple choice zener diode special purpose diodes electronic devices semiconductor electronics: materials, devices and simple circuits physics

Two capacitors each having capacitance C and breakdown voltage V are joined in series. The capacitance and the breakdown voltage of the combination will be 

  1. 2 C and 2 V

  2. C/2 and V/2

  3. 2 C and V/2

  4. C/2 and 2 V.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We know that the charge on a capacitor is 
$\begin{array}{l} Q=CV \\ C\propto \dfrac { 1 }{ V }  \\ in\, \, series\, \, combination \\ { C _{ net } }=\dfrac { { { C _{ 1 } }{ C _{ 2 } } } }{ { { C _{ 1 } }+{ C _{ 2 } } } }  \\ Hence, \\ { C _{ eq } }=\dfrac { C }{ 2 }  \\ { V _{ combination } }=2V \end{array}$

$\therefore$ option $D$ is correct.
Multiple choice physics capacitance capacitors in parallel combination of capacitors capacitors in parallel and series

A parallel plate capacitor has two square plates with equal and opposite charges. The surface charge densities on the plate are $+\sigma$ and $-\sigma$ respectively. In the region between the plates the magnitude of electric field is:

  1. $\dfrac { \sigma }{ 2{ \varepsilon } _{ 0 } } $
  2. $\dfrac { \sigma }{ { \varepsilon } _{ 0 } } $

  3. 0

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The magnitude of the electric field due to a charged plate is given as:

$\vec E=\dfrac{\sigma}{2\varepsilon _0}$

The charge density on the upper plate is positive whereas on the lower plate it is negative.

Therefore, the electric field in the region between the two plates is the difference of the two fields.
It is given as:
$E _{net}=\dfrac{\sigma}{2\varepsilon _0}-\dfrac{-\sigma}{2\varepsilon _0}$

$E _{net}=\dfrac{\sigma}{\varepsilon _0}$

Multiple choice physics capacitance capacitors in parallel combination of capacitors capacitors in parallel and series

A capacitors of $2 \mu F$ is required is an electric circuit across a potential difference of 1.0kv. A large number of $1 \mu F$ capacitors are available which can with stand a potential difference of not more than 300V The minimum number of capacitors required to achieve this is:

  1. 24

  2. 32

  3. 2

  4. 16
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

To withstand 1000V using 300V capacitors, we need at least 4 in series (4 * 300 = 1200V). To get 2uF total from 1uF capacitors, we need 2 parallel branches of 4 capacitors each, totaling 8 capacitors. However, the calculation for 1000V/300V requires 4 in series, and to get 2uF from 1uF (where each series branch is 0.25uF), we need 8 branches. 8 * 4 = 32.

Multiple choice physics capacitance capacitors in parallel combination of capacitors capacitors in parallel and series

Two capacitors were charged to potentials 80 and 30 V. Then they connected in parallel. The potential difference across both condensers is 60 V. The ratio of the capacitances of the capacitors is

  1. 3 : 2

  2. 1 : 3

  3. 1 : 4

  4. 2 : 3
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using charge conservation, Q1 + Q2 = (C1 + C2) * V_common. C1(80) + C2(30) = (C1 + C2) * 60. 80C1 + 30C2 = 60C1 + 60C2. 20C1 = 30C2. C1/C2 = 30/20 = 3/2.

Multiple choice physics capacitance capacitors in parallel combination of capacitors capacitors in parallel and series

The plates of a parallel plate capacitor are $4$cm apart, the first plate is at $300$V and the second plate at $-100$V. The voltage at $3$cm from the second plate is?

  1. $200$V
  2. $400$V
  3. $250$V
  4. $500$V
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The potential varies linearly between plates. Plate 1 is at 300V (x=0), Plate 2 is at -100V (x=4cm). The potential at distance x from Plate 1 is V(x) = 300 - (300 - (-100))/4 * x = 300 - 100x. At 3cm from Plate 2 (which is 1cm from Plate 1), V = 300 - 100(1) = 200V.

Multiple choice physics capacitance capacitors in parallel combination of capacitors capacitors in parallel and series

Find the total capacitance for three capacitors of $10$f,$15$f and $35$f in parallel with each other?

  1. $20f$
  2. $50f$
  3. $60f$
  4. $10f$
  5. $5f$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given :    $C _1 = 10$ f                  $C _2 = 15$  f                    $C _3 = 35$  f

Equivalent capacitance for parallel combination          $C _{eq} = C _1 + C _2 + C _3$
$\therefore$   $C _{eq} = 10 + 15 + 35  = 60$  $f$

Multiple choice physics capacitance capacitors in parallel combination of capacitors capacitors in parallel and series

Two capacitors of capacity $C _1$ and $C _2$ are connected in parallel, then the equivalent capacity is:

  1. $C _1+C _2$
  2. $C _1C _2/(C _1+C _2)$
  3. $C _1/C _2$
  4. $C _2/C _1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$C _1, C _2$ are connected in parallel then equivalent capacitance is calculated as
$V=V _1=V _2$.....(1)
$q=q _1+q _2$
$\therefore CV=C _1V _1+C _2V _2$
From (1) $C=C _1+C _2$

Multiple choice physics capacitance capacitors in parallel combination of capacitors capacitors in parallel and series

Capacity of a parallel plate capacitor is $2\mu F$. The two plates of the capacitor are given $400\mu C$ and $-200\mu C$charges respectively. The potential difference between the plates is 

  1. $100\ V$
  2. $200\ V$
  3. $300\ V$
  4. $150\ V$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The potential difference between plates is V = (Q1 - Q2) / (2C) is incorrect. The potential difference is determined by the charge on the inner surfaces. For a capacitor with charges Q1 and Q2, the charge on the inner faces is (Q1 - Q2)/2. Here, (400 - (-200))/2 = 300uC. V = Q/C = 300uC / 2uF = 150V.

Multiple choice physics capacitance capacitors in parallel combination of capacitors capacitors in parallel and series

A parallel plate capacitor consist of two circular plates each of radius 2 cm, separated by a distance of 0.1 mm. If voltage across the plates is varying at the rate of $5 \times {10^{13}}V{s^{ - 1}}$ , then the value of displacement current is:

  1. $5.50A$
  2. $ 5.56 \times 10^2 A $
  3. $ 5.56 \times 10^3 A $
  4. $ 2.28 \times 10^4 A $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Displacement current Id = e0 * d(Phi_E)/dt = e0 * A * dE/dt = e0 * A * (1/d) * dV/dt = (e0 * A / d) * dV/dt = C * dV/dt. C = e0 * pi * r^2 / d. C = (8.85e-12 * 3.14 * 0.02^2) / 0.0001 = 1.11e-10 F. Id = 1.11e-10 * 5e13 = 5.55 A.

Multiple choice physics capacitance capacitors in parallel combination of capacitors capacitors in parallel and series

When $n$ identical capacitors are connected in series their effective capacity is $C _s$ and when they are connected in parallel their effective capacity is $C _p$. The relation between $C _p$ and $C _s$ is:

  1. $C _p = n \,C _s$
  2. $C _p = \dfrac{C _s}{n}$
  3. $C _p = n^2 \,C _s$
  4. $C _p = \dfrac{C _s}{n^2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For n identical capacitors of capacity C, Cs = C/n and Cp = nC. Therefore, Cp = n * (n * Cs) = n^2 * Cs.