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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

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
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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. <span>$320 \mu \, C; 60 V.$</span>
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.