Tag: physics

Questions Related to physics

A capacitor contains two square plates with side lengths $5.0$ cm. The plates are separated by $2.0$ mm. Dry air fills the space between the plates. Dry air has a dielectric constant of $1.00$ and experiences dielectric breakdown when the electric field exceeds $3.0 \times  10^4$ V/cm.
What is the magnitude of charge that can be stored on each plate before the capacitor exceeds its breakdown limit and sends a spark between the plates?

parallel plate capacitor electrostatic potential and capacitance electrostatics physics

  1. $6.6 \times  10^{-8}C$

  2. $6.6 \times 10^{-5}C$

  3. $3.3 \times  10^{-7}C$

  4. $3.3 \times  10^{-84}C$

  5. $8.1 \times  10^{-2}C$

Show answer
Correct Option: A
Explanation:

Given :   $d = 2$  mm             $E = 3.0 \times 10^4$ $V/cm = 3.0 \times 10^6$  $V/m$         $l = 5.0 cm  = 0.05$ m

Area of each plate       $A = l^2 = (0.05)^2  = 25 \times 10^{-4}$  $m^2$
Capacitance of the capacitor         $C = \dfrac{A\epsilon _o }{d}$

Potential difference between the plates         $V = Ed$
$\therefore$   Charge on each plate      $Q = CV = A\epsilon _o E$
$\implies$   $Q = 25\times 10^{-4} \times 8.85 \times 10^{-12} \times 3.0 \times 10^6  = 6.6 \times 10^{-8}$  C

An air-gap parallel plate capacitor is fully charged by a battery.
What combination of two measurements will allow someone to calculate the magnitude of the electric field in between the capacitor plates?

parallel plate capacitor electrostatic potential and capacitance electrostatics physics

  1. The potential difference of the battery and the area of the plates.

  2. The charge on the plates and the distance between the plates.

  3. The charge on the plates and the area of the plates.

  4. The area of the plates and the distance between the plates.

  5. More than two measurements are needed to calculate the electric field in between the capacitor plates.

Show answer
Correct Option: C
Explanation:

Capacitance of the parallel plate capacitor       $C = \dfrac{A\epsilon _o}{d}$

Using     $Q = CV$               $\implies V = \dfrac{Q}{C}$
 Electric field between the plates      $E = \dfrac{V}{d}$
$\therefore$   $E =\dfrac{Q}{C d} = \dfrac{Q}{\frac{A\epsilon _o}{d} \times d} = \dfrac{Q}{A\epsilon _o}$
Thus option C is correct.

Two capacitors of $10\ pF$ and $20\ pF$ are connected to $200\ V$ and $100\ V$ sources respectively. If they are connected in parallel by the wire, what is the common potential of the capacitors?

parallel plate capacitor electrostatic potential and capacitance electrostatics physics

  1. $133.3\ Volt$

  2. $150\ Volt$

  3. $300\ Volt$

  4. $400\ Volt$

Show answer
Correct Option: A
Explanation:

Total charge on the two capacitors is:

$Q = Q _1 +Q _2 = C _1V _1+C _2V _2 $
$Q = 10 \times 10^{-9} \times 200 + 20 \times 10^{-9} \times 100$
$Q = 4\ \mu C$

Net capacitance of two capacitors in parallel is:
$C = C _1+C _2$
$C = 30\ pF$

Common potential of the parallel combination of capacitors is:
$V = \cfrac{Q}{C}$
$V = \cfrac{4\times 10^{-6}}{30 \times 10^{-9}} = 133.3\ V$

A thunder cloud and the earth's surface may be regarded as a pair of charged parallel plates separated by a distance $h$ and the capacitance of the system is $C$. When a flash of mean current '$i$' occurs for a time duration '$t$', the electric field strength between the cloud and earth is:

parallel plate capacitor electrostatic potential and capacitance electrostatics physics

  1. $\dfrac { it }{ C } $

  2. $Cit$

  3. $\dfrac { it }{ Ch } $

  4. $\dfrac { Cit }{ h } $

Show answer
Correct Option: C
Explanation:

Total charge accumulation in time duration $t$ is:

$Q = it$

From definition of capacitance, voltage induced due to this charge is:
$V = \cfrac{Q}{C}$
$V = \cfrac{it}{C}$

Electric field is defined as negative of gradient of potential. Hence,
$E = -\cfrac{dV}{dr}$
$\left| E \right| = \cfrac{V}{h}$
$\left| E \right| = \cfrac{it}{Ch}$

You measure the capacitor and inductor voltages in a driven RLC circuit, and find 10V for the rms capacitor voltage and 15V for the rms inductor voltage.

parallel plate capacitor electrostatic potential and capacitance electrostatics physics

  1. $\omega = \omega _{res}$

  2. $\omega < \omega _{res}$

  3. $\omega > \omega _{res}$

  4. Can't be said

Show answer
Correct Option: C
Explanation:
A/c to ques $V _{c}=10\ V$ and $V _{2}=15\ V$ 
Let frequency be $W$ capacitance be $C$ and inductance be $L$
$\Rightarrow iX _{C}=10$ and $iX _{L}=15$
$\Rightarrow \dfrac{i}{WL}=10$ and $i(WL)=15$
Dividing both
$\dfrac{\dfrac{i}{WC}}{i VWL}=\dfrac{10}{15}$
$\Rightarrow W^{2}=\dfrac{3/2}{LC}$
$W=\dfrac{\sqrt{1.5}}{\sqrt{LC}}$
we know that $W _{resonance}=\dfrac{1}{\sqrt{LC}} .... (2)$
Clearly from $(1)$ and $(2)$ $W>W _{res}(C)$

The frequency for which $5\mu F$ capacitor has a reactance of $10,000 \Omega$ is 

parallel plate capacitor electrostatic potential and capacitance electrostatics physics

  1. $(100/\pi )$ cycles per second

  2. $(10/\pi )$ cycles per second

  3. $200$ cycles per second

  4. $5,000$ cycles per second

Show answer
Correct Option: B
Explanation:

$\ = 5 \times {10^{ - 6}}f$

${x _c} = 10000\Omega $
${x _c} = \dfrac{1}{{{W _C}}}$
$ \Rightarrow w = \dfrac{1}{{c \times {x _c}}} = \dfrac{1}{{5 \times {{10}^{ - 6}} \times 10000}}$
$ = \dfrac{{{{10}^6}}}{{5 \times {{10}^4}}} = \dfrac{{100}}{5} = 20\,\,rad/\sec $
$ \Rightarrow 2\pi f = 20\,\,rad/\sec $
$ \Rightarrow f = \dfrac{{10}}{\pi }\,\,cycles/\sec $
Hence,
option $(B)$ is correct answer.

Two point charges $17.7 \mu c$ and $-17,7 \mu c$ separated by a very small distance, are kept inside a large hollow metallic sphere. Electric flux emnating through the sphere is :

parallel plate capacitor electrostatic potential and capacitance electrostatics physics

  1. $ 2 \times 10^6$Vm

  2. $- 2 \times 10^6$Vm

  3. Zero

  4. $ 4 \times 10^6$Vm

Show answer
Correct Option: C
Explanation:

Since net charge is zero

$\therefore$ Net flux is zero.
Hence,
option $C$ is correct answer.

A parallel plate capacitor has an electric field of $105$V /m between the plates .If the charge on one of the capacitor plate is 1$\mu$C,then the magnitude of the force on each capacitor plate is :

parallel plate capacitor electrostatic potential and capacitance electrostatics physics
    1. 1 N
    1. 05 N
    1. 5 N
    1. 01 N
Show answer
Correct Option: B
Explanation:

Force of attraction between the plates capacitor 

$F=\cfrac {1}{2} Q({\dfrac{Q}{A\epsilon _0} })$

=$\cfrac{1}{2} QE$

$\dfrac{1}{2}\times 10^{-6}\times 10^5$

$\dfrac{1}{20}$

$=0.05 N$

In 1909, Robert Millikan was the first to find the charge of an electron in his now-famous oil-drop experiment. In that experiment, tiny oil drops were sprayed into a uniform electric field between a horizontal pair of oppositely charged plates.The drops were observed with a magnifying eyepiece, and the electric field was adjusted so that the upward force on some negatively charged oil drops was just sufficient to balance the downward force of gravity. That is, when suspended, upward force qE just equaled mg. Millikan accurately measured the charges on many oil drops and found the values to be whole number multiples of $1.6  \times 10^{-19} C$ the charge of the electron. For this, he won the Nobel prize. Extra electrons on this particular oil drop (given the presently known charge of the electron) are :

parallel plate capacitor electrostatic potential and capacitance electrostatics physics

  1. $4$

  2. $3$

  3. $5$

  4. $8$

Show answer
Correct Option: A
Explanation:

Given that electrostatic force is just balancing gravitational force.
$qE = mg$   

$q \times 1.68 \times 10^5 = 1.08 \times 10^{-14} g$
Charge of drop, $q = 6.40 \times 10^{-19} C$
Charge of an electron $=1.6\times 10^{-19}$
Let the no. of electrons on the drop is $n$.
Then $ne=6.40 \times 10^{-19} C$
$n=\dfrac{6.40 \times 10^{-19} }{1.6\times 10^{-19}}=4$

Energy available from the oceans is:

physics sources of energy wave energy tidal energy wave power and tidal power

  1. OTE

  2. sea-waves energy

  3. tidal energy

  4. all of the above

Show answer
Correct Option: D
Explanation:

Renewable energy obtained from the ocean is together termed as ocean energy. Thus, ocean energy can be classified into various categories such as ocean thermal energy, tidal energy and wave energy.