Tag: electrostatics

Questions Related to electrostatics

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. $5q _0 \, 2\pi \, sin2\pi \nu t$


Correct Option: B
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.

The capacitance of a capacitor does not depend on

  1. the separation between the plates

  2. the size of the plates

  3. the charges on the plates

  4. none of the above


Correct Option: D
Explanation:

$C=\cfrac{Q}{V}=\cfrac{\epsilon _oA}{d}$

Capacitance depends on charge, size and separation between the plates.

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
    $


Correct Option: C
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 $Q _1=8 \ \mu C ,Q _2=16 \ \mu C , Q _3=24 \ \mu C$

A capacitor of $40\ \mu F$ charged upto $1000\ V$ is joined in parallel to another capacitor of $20\ \mu F$ charged upto $400\ V$. What is the common potential difference between the two ends of their connection ?

  1. $900\ V$

  2. $700\ V$

  3. $850\ V$

  4. $1000\ V$


Correct Option: C

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. $\dfrac{C}{2}, 2V$


Correct Option: D
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

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. $\dfrac {0.01\ln {2}}{4 \pi \epsilon _{0}}m\ V$


Correct Option: A

If on combining two charged bodies, the current does not flow then :

  1. charge is equal on both

  2. capacitance is equal on both

  3. potential is equal on both

  4. resistance is equal on both


Correct Option: C
Explanation:

The current means the charge will flow from one region to other region due to the potential difference between two regions. So if the potential is equal on both , the current will not flow.

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


Correct Option: C
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:
$q _1 = 160 \mu C$
$q _2=240 \mu C$

$V=\dfrac{q _2}{C _2}=80 V$

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 }}
    $


Correct Option: A
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 $=(\cfrac{Q}{C _1+C _2})(\cfrac{Q}{C _1+C _2})$

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. $320 \mu \, C; 60 V.$


Correct Option: A
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.