Tag: physics

Questions Related to physics

The capacitance of a capacitor does not depend on

parallel plate capacitor electrostatic potential and capacitance electrostatics physics

  1. the separation between the plates

  2. the size of the plates

  3. the charges on the plates

  4. none of the above

Show answer
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.

parallel plate capacitor electrostatic potential and capacitance electrostatics physics

  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
    $

Show answer
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 ?

parallel plate capacitor electrostatic potential and capacitance electrostatics physics

  1. $900\ V$

  2. $700\ V$

  3. $850\ V$

  4. $1000\ V$

Show answer
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:-

parallel plate capacitor electrostatic potential and capacitance electrostatics physics

  1. $2C, 2V$

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

  3. $2C, V$

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

Show answer
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 ?

parallel plate capacitor electrostatic potential and capacitance electrostatics physics

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

Show answer
Correct Option: A

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

parallel plate capacitor electrostatic potential and capacitance electrostatics physics

  1. charge is equal on both

  2. capacitance is equal on both

  3. potential is equal on both

  4. resistance is equal on both

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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 :

parallel plate capacitor electrostatic potential and capacitance electrostatics physics

  1. 200 V

  2. 100 V

  3. 80 V

  4. 40 V

Show answer
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.

parallel plate capacitor electrostatic potential and capacitance electrostatics physics

  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 }}
    $

Show answer
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.

parallel plate capacitor electrostatic potential and capacitance electrostatics physics

  1. $120 \mu \, C; 20 V.$

  2. $200 \mu \, C; 20 V.$

  3. $320 \mu \, C; 80 V.$

  4. $320 \mu \, C; 60 V.$

Show answer
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.

A glass slab is put within the plates of a charged parallel plate condenser. Which of the following quantities does not change?

parallel plate capacitor electrostatic potential and capacitance electrostatics physics

  1. energy of the condenser

  2. capacity

  3. intensity of electric field

  4. charge

Show answer
Correct Option: B