Tag: capacitors in parallel and series

Questions Related to capacitors in parallel and series

$4\ \mu F$ and $6\ \mu F$ capacitors are joined in series and $500\ v$ are applied between the outer plates of the system. What is the charge on each plate ?

  1. $1\cdot 2\times 10^{3}\ C$

  2. $6\cdot 0\times 10^{3}\ C$

  3. $5\cdot 0\times 10^{-3}\ C$

  4. $2\cdot 0\times 10^{-3}\ C$


Correct Option: A
Explanation:

Since the capacities are connected in series

$\dfrac{1}{C _{eq}}=\dfrac{1}{C _1}+\dfrac{1}{C _2}$
$\dfrac{1}{C _{eq}}=\dfrac{1}{4\mu}+\dfrac{1}{6\mu}$
$C _{eq}=\dfrac{12\mu}{5}$
$C=\dfrac{Q}{V}$
$\dfrac{12\mu}{5}=\dfrac{Q}{500}$
$Q=1.2\times 10^{3}$

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


Correct Option: B
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:
$E _{net}=\dfrac{\sigma}{2\varepsilon _0}-\dfrac{-\sigma}{2\varepsilon _0}$

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

If the voltage across the parallel combination in increases which capacitor will undergo breakdown first?

  1. $C$

  2. $2C$

  3. Both at same moment

  4. None of these


Correct Option: C

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


Correct Option: B

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


Correct Option: A

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


Correct Option: A

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. $5f$


Correct Option: C
Explanation:

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

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

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. $C _2/C _1$


Correct Option: A
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) $C=C _1+C _2$

A parallel plate capacitor is connected to a battery and the distance between the plates is decreased then which quantity is same for the parallel plate capacitor

  1. stored energy

  2. capacitance

  3. intensity of electric field

  4. potential difference


Correct Option: A

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. $150\ V$


Correct Option: D