Physics

Capacitors and Capacitance

186 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.

Series and parallel capacitorsParallel plate capacitorsCapacitive reactance formulasDielectrics and permittivitySpherical capacitors

Capacitors and Capacitance Questions

Multiple choice physics option a: relativity maxwell's equations the nature of light introduction to electromagnetic waves

A parallel plate capacitor having plate area A and plate separation $d$ is connected to a battery of emf $\varepsilon$ and internal resistance $R$ at $t=0$. Consider a plane surface of area $\dfrac{A}{2}$, parallel to the plates and situated symmetrically between them. Find the displacement current through this surface as a function of time?

  1. $\dfrac {-\varepsilon}{2R} \ \ \ e^{\dfrac{-td}{\varepsilon AR}}$
  2. $\dfrac {2\varepsilon}{R} \ \ \ e^{\dfrac{-td}{\varepsilon AR}}$
  3. $\dfrac {5\varepsilon}{2R} \ \ \ e^{\dfrac{-td}{4 \varepsilon AR}}$
  4. $\dfrac {\varepsilon}{2R} \ \ \ e^{\dfrac{-td}{4\pi \varepsilon AR}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The displacement current in a charging capacitor is equal to the conduction current in the circuit. The current in an RC circuit is I(t) = (epsilon/R) * e^(-t/RC). Since the surface area A/2 is parallel to the plates, it intercepts half the electric flux, leading to a factor of 1/2.

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

Two capacitors of $1\mu F$ and $2\mu F$ are connected in series and this combination is changed upto a potential difference of $120$ volt. What will be the potential difference across $1 \mu F$ capacitor:

  1. $40 volt$
  2. $60 volt$
  3. $80 volt$
  4. $120 volt$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given, $c _1=1\mu f,c _2=2\mu f,PD=120v$

$C _{eq}=\dfrac{c _1c _2}{c _1+c _2}=\dfrac{2\times1}{2+1}=\dfrac{2}{3}\mu f$

We know,  $Q=cv$ Where Q is the charge, C is the capacitance of the capacitor and v is the potential difference.

Now, $Q _{net}$ in circuit is equivalent capacitance of capacitors attached in the circuits is multiplied by PD

$Q _{net}=12\times\dfrac{2}{3}=80$

$Q=cv=1\mu fv=80\Rightarrow v=80v$
Multiple choice physics capacitance capacitors in series combination of capacitors capacitors in parallel and series

Two capacitors of caacity ${ C } _{ 1 }$ and ${ C } _{ 2}$ are connected in series and potential difference V is applied across it. Then the potential difference across${ C } _{ 1 }$ will be 

  1. $V\frac { { C } _{ 2 } }{ { C } _{ 1 } } $
  2. $V\frac { { C } _{ 1 }+{ C } _{ 2 } }{ { C } _{ 1 } } $
  3. $V\frac { { C } _{ 2 } }{ { C } _{ 1 }+{ C } _{ 2 } } $
  4. $V\frac { { C } _{ 1 } }{ { C } _{ 1 }+{ C } _{ 2 } } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For two capacitors connected in series, the charge Q is the same on both. The potential difference across C1 is V1 = Q / C1. Since equivalent capacitance is C_eq = (C1 * C2) / (C1 + C2) and total charge is Q = C_eq * V, substituting gives V1 = V * (C2 / (C1 + C2)).

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

For capacitors in the series combination, the total capacitance C is given by

  1. $C=(\cfrac{1}{C _1}+\cfrac{1}{C _2} + ......)$
  2. $C = C _{1} + C _{2} +$ .......
  3. $\cfrac{1}{C}=(\cfrac{1}{C _1}+\cfrac{1}{C _2}+.....)$
  4. $\cfrac{1}{C} = C _{1} + C _{2} +$ ........
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

When in series, the reciprocal of the net capacitance is equal to the sum of reciprocal of individual capacitances.

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

A series combination of two capacitances of value $0.1\ mu F$ and $1\mu F$ is connected with a source of voltage $500\ volts$. The potential difference in volts across the capacitor of value $0.1\ muF$ will be :

  1. $50$
  2. $500$
  3. $45.5$
  4. $454.5$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given,

Capacitance, ${{C} _{1}}=0.1\,\mu F\,\,and\,\,{{C} _{2}}=1\,\mu F$

In series charge is equal

$ Q={{C} _{1}}{{V} _{1}}={{C} _{2}}{{V} _{2}} $

$ {{V} _{2}}=\dfrac{{{C} _{1}}{{V} _{1}}}{{{C} _{2}}} $

In series total potential difference is sum of all paternal difference

$ V={{V} _{1}}+{{V} _{2}} $

$ V={{V} _{1}}+\dfrac{{{C} _{1}}{{V} _{1}}}{{{C} _{2}}}={{V} _{1}}\left( \dfrac{{{C} _{2}}+{{C} _{1}}}{{{C} _{2}}} \right) $

$ {{V} _{1}}=\dfrac{{{C} _{2}}V}{{{C} _{2}}+{{C} _{1}}}=\dfrac{1\times 500}{1+0.1}=454.54\,V $

Hence, Potential difference across $0.1\,\mu F\,\,\,is\,\,\,454.5\,V$ 

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

Two parallel plate capacitors are connected in series. Each capacitor has a plate area A and a separation d between the plates. The dielectric constant of the medium between their plates are 2 and 4 . The separation between the plates of a single air capacitors of plate area A which effectively replaces the combination is:

  1. 2d/3

  2. 3d/2

  3. 3d/4

  4. 8d/5
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Two capacitors of area A and separation d with dielectric constants 2 and 4 are in series. Their capacitances are C1 = 2A e0 / d and C2 = 4A e0 / d. The equivalent capacitance C_s satisfies 1/C_s = 1/C1 + 1/C2 = d/(2A e0) + d/(4A e0) = 3d/(4A e0), so C_s = 4A e0 / (3d). Equating this to a single air capacitor C = A e0 / d_eff gives d_eff = 3d/4.

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

A capacitor comprises of two parallel circular plates. Diameter of each of plates is equal to $6 cm$. If capacitance of above system is equivalent to capacitance of sphere, whose diameter is equal to $200 cm$. Distance between two plates will be:-

  1. $2.25 \times 10^{-4} m$
  2. $4.5 \times 10^{-4} m$
  3. $6.75 \times 10^{-4} m$
  4. $9 \times 10^{-4} m$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The capacitance of a parallel plate capacitor is C = epsilon_0 * A / d = epsilon_0 * pi * r^2 / d, where r = 3 cm = 0.03 m. The capacitance of a conducting sphere of radius R = 100 cm = 1 m is C_sphere = 4 * pi * epsilon_0 * R. Equating the two: pi * epsilon_0 * (0.03)^2 / d = 4 * pi * epsilon_0 * (1). Solving for d gives d = 0.0009 / 4 = 2.25 * 10^-4 m.

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

Two identical capacitors are connected in series with a source of potential V. If Q is the charge on one of the capacitors, the capacitance of each capacitor is: 

  1. Q/2V

  2. 2Q/V

  3. Q/V

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

In series connection charge on each capacitor would be constant also equivalent capacitance in series $c'=\dfrac{C}{2}$ [ following $\dfrac{1}{c'}=\dfrac{1}{c _1}+\dfrac{1}{c _2}$] and voltage $V$ is applied across it so,from capacitive law,$Q=c'v \Rightarrow Q=\dfrac{CV}{2} \Rightarrow C=\dfrac{2Q}{V}$

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

Two capacitors of $4\ \mu F$and $2\ \mu F$ are connected in series with the battery. If total potential difference across the two capacitors is $200$ volts then the ratio  of potential difference across one capacitor to another is

  1. $1:2$
  2. $2:1$
  3. $1:4$
  4. $4:1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In series, charge Q is constant. V = Q/C. Therefore, V1/V2 = C2/C1. With C1=4uF and C2=2uF, V1/V2 = 2/4 = 1/2.