MCQ Questions for Class 12 Medical Physics Electrostatic Potential And Capacitance Quiz 8

Equipotential surfaces are

0%
Surfaces that are perpendicular to gravitational fields
0%
Surfaces that have same mass on top of it
0%
Surface that have same radius of curvature
0%
Surfaces that have same gravitational field on it

Explanation

Equipotential surfaces are surfaces which have the same value of potential throughout their surface.
For a gravitatiomal field of $E=\dfrac{GM}{r^2}$ ,the potential is $v=-\dfrac{GM}{r}$
So,at a particular 'r',the potential remains constant at a value of $-\dfrac{GM}{r}$
So equipotential surfaces are circles for the filed of $E=\dfrac{GM}{r^2}$
Hence,the field is perpendicular to equipotential surface.

In the figure a capacitor of capacitance

$2\mu F$ as is connected to a cell of eml 20 volt. The plates of the capacitors are drawn apart slowly to double the distance between them, The work done by the external agent on the plates is:

0%
$-200 \mu J$
0%
$200 \mu J$
0%
$400 \mu J$
0%
$-400 \mu J$

Explanation

When the plate separation is double, the charge remains constant and the capacitance is halved.The Work done in separating the plate is the difference in energy stored.

$\cfrac { { Q }^{ 2 } }{ 2\left( \cfrac { C }{ 2 } \right) } -\cfrac { { Q }^{ 2 } }{ 2C } =-\cfrac { { Q }^{ 2 } }{ 2C }$

$-\cfrac{\left(CV\right)^2}{2C}$

$-\cfrac{CV^2}{2}$

$=\cfrac{1}{2}\times 2\times \left(20\right)^2$

$=-400 \mu J$

The electrostatic potential energy of two point charges,

$1$

$\mu$ C each, placed

$1$ meter apart in air is?

0%
$9\times 10^3$ J
0%
$9\times 10^9$ J
0%
$9\times 10^{-3}$ J
0%
$9\times 10^{-3}$ eV

Explanation

$\textbf{Step 1: Formula of Potential energy}$

Electrostatic potential energy is given by

$U$ between two charges placed at a separtion

$r$ is given by

$\Rightarrow\ \ U=\dfrac { k{ q }_{ 1 }{ q }_{ 2 } }{ r}$

$\textbf{Step 2: Calculations}$

Putting values,

$k=9\times { 10 }^{ 9 }\dfrac { N.{ m }^{ 2 } }{ { C }^{ 2 } } ,$

${ q }_{ 2 }={ q }_{ 1 }=1\mu C=1\times { 10 }^{ -6 }C,$

$r=1m$

$\\ \Rightarrow U\\$

$=\dfrac { 9\times { 10 }^{ 9 }\times 1\times { 10 }^{ -6 }\times 1\times { 10 }^{ -6 } }{ 1 } =9\times { 10 }^{ -3 }J\\$

Hence, correct option is

$C$ .

Charges

$1\mu C$ are placed at each of the four corners of a square of side

$2\sqrt {2}m$ . The potential at the point of intersection of the diagonals is _____

$(K = 9\times 10^{9} SI\ unit)$ .

0%
$18\times 10^{3}V$
0%
$1800\ V$
0%
$18\sqrt {2}\times 10^{3}V$
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None of these

Explanation

Key Concept Here, we have to find the electric potential due to multiple of charges, so, we have to apply superposition principle to calculate net electric potential.
Length of the diagonals

$= \sqrt {2}\times$ side length

$= \sqrt {2} \times 2\sqrt {2} = 4\ m$

$\therefore$ Length of semi-diagonals

$= \dfrac {4}{2} = 2m$ .
Now, potential due to charges at point

$O$

$V_{0} = \dfrac {!}{4\pi \epsilon_{0}} \left (\dfrac {1\times 10^{-6}}{2}\right ) \times 4$

$= 9\times 10^{9}\times 10^{-6} \times 2 = 18\times 10^{3}V$ .

The potential energy of system of two equal negative point charges of

$2\mu C$ each held 1 m apart in air is (

$k = 9 \times 10^9\, SI \,unit$ )

0%
$36 \,J$
0%
$3.6 \times 10^{-3}\, J$
0%
$3.6 \,J$
0%
$3.6 \times 10^{-2}\, J$

Explanation

The potential energy of system is

$U=k\dfrac{q_1q_2}{r}$

$U=k\dfrac{q^2}{r}$

$[q_1=q_2=q]$
or

$U= 9\times 10^9 \times \dfrac{(2\times 10^{-6})^2}{1}$
or

$U=9\times 10^9\times 4 \times 10^{-12}$
or

$U=3.6 \times 10^{-2}\, J$

A hollow metal sphere of radius

$10$ cm is charged such that the potential on its surface becomes

$80$ V. The potential at the centre of the sphere is?

0%
$80$ V
0%
$800$ V
0%
$8$ V
0%
Zero

Explanation

The potential at the centre of the sphere is

$80$ V because it remains same at each point under the metallic hollow sphere.

A hollow metal ball carrying an electric charge produces no electric field at points.

0%
Outside the sphere
0%
On its surface
0%
Inside the sphere
0%
At a distance more than twice

Explanation

Electric field due to a charge carrying hollow metallic sphere is given by

$E_r = 0$ for

$r< R$
where

$R$ is the radius of sphere.

$E_r = \dfrac{KQ}{r^2}$ for

$r\geq R$
So, electric field inside the hollow metallic sphere is zero but non zero outside it.

A uniformly charged fixed ring Q of radius R has a small charge

$-q$ on its axis at a point A ( OA

$=$ d and d

$< <$ R). When the charge was released, the distance AO was covered in time T. Then which of the following statements is/are true.

0%
For OB $=2$ d, if charge was released from B, then the distance AO would be covered in a time $T/6$ .
0%
For OB $=\sqrt{2}$ d if charge was released from B then the distance AO would be covered in a time $T/2$
0%
If Q is doubled, then T doubles
0%
If R is doubled, then T decreases

Explanation

Time period

$=4T(SHM$ and is independent of d

So, if

$OB=2d$ , coverage of AO requires

$\cfrac{1}{12}th$ of time period

$=\cfrac{4T}{12}=\cfrac{T}{3}$

and if

$OB=\sqrt2 d$ , coverage of AO requires

$\cfrac{1}{8}th$ of time period

$\cfrac{4T}{8}=\cfrac{T}{2}$

Time period is inversely proportion to

$\omega$ and

$\omega^2=\cfrac{Q_9}{Q^3}$ (for small

$d=\cfrac{Qd_cd }{4\pi E_0(d^2+R^2)^{3/2}}\\ \quad=\cfrac{Q_9d}{4\pi E_0R^3})$

If Q doubles,

$\omega$ becomes

$\sqrt2$ times and

$T$ becomes

$\cfrac{1}{\sqrt2}$ times

Option B

1219149_769164_ans_3609f36d0cc54270b6b1ef54644d5ff6.png

The ratio of momentum of an electron and an alpha particle which are accelerated from rest by potential difference of 100 V is:

0%
$\sqrt{\dfrac{m_{\alpha}}{m_e}}$
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$\sqrt{\dfrac{m_e}{m_{\alpha}}}$
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$\dfrac{2m_e}{m_{\alpha}}$
0%
$\sqrt{\dfrac{m_e}{2m_{\alpha}}}$

Explanation

Mass of proton

$=1$

$u$

and, Mass of alpha particle

$=4$

$u$

Charge of proton

$=e$

And, charge of alpha particle

$=2e$

For charge moving in a uniform potential field.

$K.E=$ Electric potential energy

$\cfrac{1}{2}mv^2=qV$

$v=\sqrt{\cfrac{2qV}{m}}$

OR,

Momentum

$(P)=mv$

$\Rightarrow P=m\sqrt{\cfrac{2qV}{m}}$

$\Rightarrow P=\sqrt{m2qV}$

Momentum of proton,

$P_1=\sqrt{2qVm_1}=\sqrt{2eVm_e}$

Momentum of alpha particle

$P_2=\sqrt{2qVm_2}=\sqrt{2\times 2eVm_a}$

$\therefore \cfrac{P_1}{P_2}=\sqrt{\cfrac{2eVm_e}{4eVm_a}}$

$\therefore \cfrac{P_1}{P_2}=\sqrt{\cfrac{m_e}{2m_a}}$

A charged particle 'q' is shot with speed v towards another fixed charged particle Q. It approaches Q upto a closest distance r and then returns. If q were given a speed

$2v$ , the closest distance of approach would be.

0%
r
0%
$2r$
0%
$r/2$
0%
$r/4$

Explanation

At closest distance r its whole KE is converted into PE.

$\therefore \displaystyle\frac{1}{2}mv^2=\frac{1}{4\pi \varepsilon_0}\frac{Q\cdot q}{r}$

$\Rightarrow r=\displaystyle\frac{1}{4\pi \varepsilon_0}\frac{Q\cdot q}{mv^2}$
In next case,

$\displaystyle r'=\frac{1}{4\pi \varepsilon_0}\frac{Qq}{m(2v)^2}$

$\Rightarrow r'=\displaystyle \frac{1}{4}\left(\displaystyle\frac{1}{4\pi \varepsilon_0}\cdot \frac{Qq}{mv^2}\right)\Rightarrow r'=r/4$ .

The electric potential at a point in free space due to a charge Q coulomb is

$Q\ \times { 10 }^{ 11 }$ V. The electric field at that point is:

0%
12 $\pi { \varepsilon }_{ 0 }Q \times { 10 }^{ 22 }\ V { m }^{ -1 }$
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4 $\pi { \varepsilon }_{ 0 }Q \times { 10 }^{ 22 }\ V { m }^{ -1 }$
0%
12 $\pi { \varepsilon }_{ 0 }Q \times { 10 }^{ 20 }\ V { m }^{ -1 }$
0%
$4\pi{ \varepsilon }_{ 0 }Q \times { 10 }^{ 20 }\ V { m }^{ -1 }$

Explanation

Let point

$P$ be at distance

$x$ from charge

$(1)$

Potential

$=\dfrac {kQ}{x}=Q\times 10^{11}$

$x=k/10^{11}$

Eludic field

$=\dfrac {kQ}{x^2}=\dfrac {kQ}{k^2 /10^{22}}$

$=\dfrac {Q\times 10^{22}}{k}$

$\left (k=\dfrac {1}{4\pi \epsilon _0}\right)$

$\boxed {E=4\pi \epsilon _0 Q\times 10^{22}}$

Option

$(B)$

The area of the plates of a parallel plante capacitor is

$A$ and the gap between them is

$d$ . The gap is filled with a non-homogeneous dielectric whose dielectric constant varies with the distance

$y$ from one plate as:

$K=\lambda \sec { \left( \cfrac { \pi y }{ 2d } \right) }$ , where

$\lambda$ is a dimensionless constant. The capacitance of this capacitor is

0%
$\cfrac { \pi { \varepsilon }_{ 0 }\lambda A }{ 2d }$
0%
$\cfrac { \pi { \varepsilon }_{ 0 }\lambda A }{ d }$
0%
$\cfrac { 2\pi { \varepsilon }_{ 0 }\lambda A }{ d }$
0%
None

Explanation

$E=\cfrac{\sigma}{K\epsilon_o}$

$\cfrac{dv}{dy}=\cfrac{\sigma}{K\epsilon_o}$

$\int _{ 0 }^{ v } dv=\cfrac { \sigma }{ \lambda \epsilon _{ o } } \int _{ 0 }^{ d } \cos \left( \cfrac { \pi y }{ 2d } \right) dyV\\ =\cfrac { \sigma }{ \lambda \epsilon _{ o } } \times \cfrac { 2d }{ \pi } \left[ \sin \cfrac { \pi y }{ 2d } \right] _{ 0 }^{ d }\\ =\cfrac { \sigma }{ \lambda \epsilon _{ o } } \times \cfrac { 2d }{ \pi } \\ C=\cfrac{ Q}{V}=\cfrac{A\lambda\epsilon_o\pi}{2d}$

A capacitor is charged by a battery and then the battery is disconnected. A dielectric slab is introduced between the plates. The result is

0%
P.d between the plates increases, charge on the plate decreases
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P.d between the plates decreases, charge remains same
0%
P.d increases, charge remain constant
0%
P.d decreases, charge increases

Explanation

As dielectric slab does not affect charges:

$V=\cfrac{Q}{C}=\cfrac{Qd}{K\epsilon_oA}$

Which one of the following gives the resultant capacitor when capacitors are joined in series?

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The sum of the individual capacitors
0%
The reciprocal of the sum of the reciprocals of the individual capacitors
0%
The reciprocal of the sum of the capacitors
0%
The sum of the reciprocals of the individual capacitors

Explanation

The resultant capacitor when capacitors are joined in series is the reciprocal of the sum of the reciprocals of the indivisual capacitors.

$\cfrac{1}{c_{eq}}=$

$\cfrac{1}{c_{1}}$ +

$\cfrac{1}{c_{2}}$

A system consists of two charges 4

$\mu C$ and -3

$\mu C$ with no external field placed at (-5 cm, 0, 0) and (5 cm, 0, 0) respectively. The amount of work required to separate the two charges infinitely away from each other is

0%
1.1 J
0%
2 J
0%
2.5 J
0%
3 J

Explanation

Here

${ q }_{ 1 }=4\mu C,{ q }_{ 2 }=-3\mu C,r=10cm=0.1m$

Electrostatic potential energy

$U_i=\cfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \cfrac { { q }_{ 1 }{ q }_{ 2 } }{ r } =9\times { 10 }^{ 9 }\times \cfrac { 4\times { 10 }^{ -6 }\times \left( -3 \right) \times { 10 }^{ -6 } }{ 0.1 } =-1.1J$

$U_f=0$

Work required,

$W=U_f-U_i=0-(-1.1)=1.1$

The potential at a point due to a charge of 5 x

${ 10 }^{ -7 }$ C located 10 cm away is:

0%
3.5 x ${ 10 }^{ 5 }$ V
0%
3.5 x ${ 10 }^{ 4 }$ V
0%
4.5 x ${ 10 }^{ 4 }$ V
0%
4.5 x ${ 10 }^{ 5 }$ V

Explanation

Given:

$Q=5\times 10^{-7}C$

$k=9\times 10^9$

$x=10\ cm=\dfrac {1}{10}m$

$P=Potential =\dfrac {kQ}{x}$

$P=\dfrac {9\times 10^9 \times 5\times 10^{-7}}{1/10}$

$P=4.5\times 10^4\ V$

Which of the following statement is true about the relation between electric field and potential?

0%
Electric field in the direction in which the potential decreases steepest
0%
Magnitude of electric field is given by the change in the magnitude of potential per unit displacement normal to the equipotential surface at that point.
0%
In the region of strong electric field, equipotential surfaces are far apart.
0%
Both the statements (a) and (b) are correct.

Explanation

The magnitude of electric field is given as the change in magnitude of potential per unit displacement as:

$E = -\dfrac{dV}{dx}$

(-) sign indicates that field is directed from higher potential to lower potential and it also suggests that the electric field is in the direction in which the potential decreases steepest.

Option

$(D)$ is correct.

There is a -ve charge -

$q$ as shown in the figure. Take the potential to be zero at infinity. If

$A$ and

$B$ are two points in the charge's vicinity, then the potential at-

0%
$A$ is less than at $B$
0%
$A$ is more than that at $B$
0%
$A$ is greater in magnitude than at $B$
0%
$A$ is smaller in magnitude than at $B$

Explanation

Potential at Point

$A$ will be given by,

$V_A=\cfrac{K(-q)}{A}=\cfrac{-Kq}{A}$

Potential at Point

$B$ will be given by,

$V_B=\cfrac{K(-q)}{B}=\cfrac{-Kq}{B}$

Now, As

$A<B$ ,

So, the potential at

$A$ is greater magnitude than at

$B$ .

But, as due to negative sign convention potential is greater at

$B$ than at

$A$ .

What is the angle between electric field and equipotential surfaces ?

0%
90 always
0%
0 always
0%
0 to 90
0%
0 to 180

Explanation

$dV=\vec E \cdot \vec dr$

$\because dV=0,$ In RHS dot product will be zero

$\therefore \$ An electric field is always perpendicular to the equipotential surface at any point.

Which of the following statement is not true?

0%
Electrostatic force is a conservative force.
0%
Potential at a point is the work done per unit charge in bringing a charge from infinity to that point in an electric field.
0%
Electrostatic force is non-conservative.
0%
Potential is the ratio of work to charge.

A capacitor of capacitance

${ C }_{ 1 }$ is charged to a potential V and then connected in parallel to an uncharged capacitor of capacitance

${ C }_{ 2 }$ . The final potential difference across each capacitor will be

0%
$\dfrac { { C }_{ 1 }V }{ { C }_{ 1 }+{ C }_{ 2 } }$
0%
$\dfrac { { C }_{ 2 }V }{ { C }_{ 1 }+{ C }_{ 2 } }$
0%
$1+\dfrac { { C }_{ 2 } }{ { C }_{ 1 } }$
0%
$1-\dfrac { { C }_{ 2 } }{ { C }_{ 1 } }$

An infinite cylinder of radius

${ r }_{ 0 }$ carrying linear charge density

$\lambda$ . The equation of the equipotential surface for this cylinder is

0%
$\displaystyle r={ r }_{ 0 }{ e }^{ \pi { \varepsilon }_{ 0 }\left\lfloor V(r)+V({ r }_{ 0 }) \right\rfloor \lambda }$
0%
$\displaystyle r={ r }_{ 0 }{ e }^{ 2\pi { \varepsilon }_{ 0 }\left\lfloor V(r)-V({ r }_{ 0 }) \right\rfloor { \lambda }^{ 2 } }$
0%
$\displaystyle r={ r }_{ 0 }{ e }^{ -2\pi { \varepsilon }_{ 0 }\left\lfloor V(r)-V({ r }_{ 0 }) \right\rfloor /\lambda }$
0%
$\displaystyle r={ r }_{ 0 }{ e }^{ -2\pi { \varepsilon }_{ 0 }\left\lfloor V(r)-V({ r }_{ 0 }) \right\rfloor \lambda }$

Explanation

Gaussian surface of radius

$r$ and length

$l$ .

According to Gauss's theorem

$\displaystyle \oint { \vec { E } .\vec { ds } } =\dfrac { q }{ { \varepsilon }_{ 0 } } =\dfrac { \lambda l }{ { \varepsilon }_{ 0 } }$

$E\left( 2\pi rl \right) =\dfrac { \lambda l }{ { \varepsilon }_{ 0 } }....(i)$

$or\quad E=\dfrac { \lambda }{ 2\pi { \varepsilon }_{ 0 } } \log _{ e }{ \dfrac { { r }_{ 0 } }{ r } }$

$\therefore V(r)-V({ r }_{ 0 })=-\displaystyle\oint _{ { r }_{ 0 } }^{ r }{ \vec { E } .\vec { dl } } =\dfrac { \lambda }{ 2\pi { \varepsilon }_{ 0 } } \log _{ e }{ \dfrac { { r }_{ 0 } }{ r } }$

For an equipotential surface of given

$V(r)$

$\log _{ e }{ \dfrac { { r }_{ 0 } }{ r } } =\dfrac { 2\pi { \varepsilon }_{ 0 } }{ \lambda } \left[ V(r)-V({ r }_{ 0 }) \right]$

$\therefore r={ r }_{ 0 }{ e }^{ -2\pi { \varepsilon }_{ 0 }\left[ V(r)-V({ r }_{ 0 }) \right] \lambda }$

1028962_945387_ans_b33df26307f446cc877f457c127cb823.png

The charge on 3

$\mu F$ capacitor shown in the figure is

0%
2 $\mu C$
0%
10 $\mu C$
0%
6 $\mu C$
0%
8 $\mu C$

Explanation

Since all the capacitors are connected in series,

$\therefore$ equivalent capacitance is

$\cfrac { 1 }{ { C }_{ eq } } =\cfrac { 1 }{ 2 } +\cfrac { 1 }{ 3 } +\cfrac { 1 }{ 6 } =1\quad \quad \therefore { C }_{ eq }=1\mu F$
Charge on each capacitor is same

$q={ C }_{ eq }V=1\times 10=10\mu C\quad$ (

$\because V=10V$ )

The figure show several equipotential lines. Comparing between points

$A$ and

$\mathrm { B }$ , pick up the best possible statement

0%
the electric field has a greater magnitude at point $\mathrm { A }$ and is directed to left.
0%
the electric field has a greater magnitude at point $\mathrm { A }$ and is directed to right.
0%
the electric field has a greater magnitude at point $\mathrm { B }$ and is directed to left.
0%
the electric field has a greater magnitude at point $\mathrm { B }$ and is directed to lright.

A parallel plate capacitor of capacitance 5

$\mu F$ and plate separation 6 cm is connected to a 1 V battery and charged. A dielectric of dielectric constant 4 and thickness 4 cm is introduced between the plates of the capacitor. The additional charge that flows into the capacitor from the battery is

0%
2 $\mu C$
0%
3 $\mu C$
0%
5 $\mu C$
0%
10 $\mu C$

Explanation

Charge on capacitor plates without the dielectric is

$=CV=\left( 5\times { 10 }^{ -6 }F \right) \times 1V=5\times { 10 }^{ -6 }C=5\mu C$
The capacitance after the dielectric is introduced is

$C'=\cfrac { { \varepsilon }_{ 0 }A }{ d-\left( t-\cfrac { t }{ K } \right) } =\cfrac { { \varepsilon }_{ 0 }A/d }{ 1-\left( \cfrac { t-\cfrac { t }{ K } }{ d } \right) } =\cfrac { C }{ 1-\left( \cfrac { t-\cfrac { t }{ K } }{ d } \right) } =\cfrac { 5\mu F }{ 1-\left( \cfrac { 4cm-\cfrac { 4cm }{ 4 } }{ 6cm } \right) } =\cfrac { 5\mu F }{ 1-\left( \cfrac { 4-1 }{ 6 } \right) } =10\mu F\quad \quad$

$\therefore$ Charge on capacitor plate now will be

$Q'=C'V=10\mu F\times 1V=10\mu C$
Additional charge transferred

$=Q'-Q=10\mu C-5\mu C=5\mu C$

A parallel capacitor plate is charged and then isolated. The effect of increasing of plate separation on charge, potential, capacitance respectively are

0%
constant, decreases, decreases
0%
increase, decreases, decreases
0%
constant, decreases, increases
0%
constant, increases, decreases.

Explanation

For parallel plate capacitor

$C= \varepsilon_0 \dfrac{A}{d}$ ... 1

For charged capacitor

$Q=CV$ .. 2

When a parallel plate capacitor is charged and then isolated the

$charge \ Q$ on the capacitor remains constant

From 1, Capacitance

$C$ is inversely proportional to

$d$ (separation between the plates)

Hence on increasing plate separation i.e

$\ d$ the

$capacitance$ of the capacitor

$decreases$

From 2,

$C$ is inversely proportional to

$V$ hence potential

$V$

$increases$ as

$C$

$decreases$

Correct option is

$D$

A capacitor has some dielectric between its plates, and the capacitor is connected to a dc source. The battery is now disconnected and then the dielectric is removed, then

0%
capacitance will increase.
0%
energy stored will decrease.
0%
electric field will increase.
0%
Voltage will decrease.

Explanation

When the capacitor is connected to dc source, it gets charged. The battery is then disconnected, so no more charge can flow in. On removing dielectric, capacitance decreases.
Energy stored

$\left( u=\cfrac { { q }^{ 2 } }{ 2C } \right)$ will increase
Potential

$\left( V=\cfrac { q }{ C } \right)$ will also increase
Electric field

$\left( E=\cfrac { V }{ D } \right)$ will increase

In a parallel plate capacitor, the capacity increases if:

0%
area of the plate is decreased
0%
distance between the plates is increases
0%
area of the plate is increased
0%
dielectric constant decreases.

Explanation

Hint:- Check the dependence of capacitance on certain quantities.

Explanation:-

$\bullet$ In a parallel plate capacitor, the capacitance is $C =$ $\dfrac{k{{\varepsilon }_{0}}A}{d}$ ;

$\bullet$ Where, k is the dielectric constant, ${{\varepsilon }_{0}}$ is the permittivity constant, A is the area of the conduction and d is the distance between plates.

$\bullet$ From here we can see C is directly proportional to k, $\epsilon_o$ ,A and inversely proportional to d.

⸫ So capacitance will increase with increase in Area.

$\textbf{Hence, the correct option is (C)}$

Two identical capacitors are joined in parallel, charged to a potential V, separated and then connected in series, the positive plate of one is connected to the negative of the other. Which of the following is true?

0%
The charges on the free plated connected together are destroyed.
0%
The energy stored in the system increases.
0%
The potential difference between the free plates is 2 V.
0%
The potential difference remains constant.

Explanation

When the two capacitors charged to same potential are connected in series, then total potential difference

$V'={ V }_{ 1 }+{ V }_{ 2 }=V+V=2V$
1071551_945105_ans_fc25ce22d6994235beed18d4429276b7.PNG

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:

0%
$\dfrac { \sigma }{ 2{ \varepsilon }_{ 0 } }$
0%
$\dfrac { \sigma }{ { \varepsilon }_{ 0 } }$
0%
0
0%
none of these

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:

$E_{net}=\dfrac{\sigma}{2\varepsilon_0}-\dfrac{-\sigma}{2\varepsilon_0}$

$E_{net}=\dfrac{\sigma}{\varepsilon_0}$

A parallel plate air capacitor is charged to a potential difference of V volts. After disconnecting the charging battery the distance between the plates of the capacitor is increased using an insulating handle. As a result the potential difference between the plates:

0%
increases
0%
decreases
0%
does not change
0%
becomes zero

Explanation

Capacitance

$C=\varepsilon\dfrac{A}{d}$ ,

$A-area\ of\ the\ capacitor\ ; d-distance\ between\ the\ plates$

For a charged capacitor

$Q=CV$

As the distance between the plates of the capacitor is increased the capacitance

$C$ decreases

$\ \ \ (\because C \propto \dfrac{1}{d})$

As capacitance decreases the potential difference between the plates

$V \ increases$

A capacitor consists of two metal plates each

$10{\text{ }}cm$ by

$20{\text{ }}cm;$ they are separated by a

$2.0{\text{ }}mm$ thick insulator with dielectric constant

$4.1$ and dielectric strength 6.0107 V/m. What is the capacitance in

$pF\left( {{{10}^{ - 12}}F} \right)?$

0%
75
0%
100
0%
240
0%
360

Explanation

$C = \cfrac{k\epsilon_0A}{d}$

$C =\cfrac{4.1\times 8.85\times 10^{-12}\times 10\times 20\times 10^{-4}}{2\times 10^{-3}}$

$C = 361\approx 360 pF$

The electric potential at a point in free space due to a charge

$Q$ coulomb is

$Q\times { 10 }^{ 11 }V$ . The electric field at that point is

0%
$12\pi { \varepsilon }_{ 0 }Q\times { 10 }^{ 22 }V{ m }^{ -1 }$
0%
$4\pi { \varepsilon }_{ 0 }Q\times { 10 }^{ 22 }V{ m }^{ -1 }$
0%
$12\pi { \varepsilon }_{ 0 }Q\times { 10 }^{ 20 }V{ m }^{ -1 }$
0%
$4\pi { \varepsilon }_{ 0 }Q\times { 10 }^{ 20 }V{ m }^{ -1 }$

Explanation

Here

$V=\cfrac { Q }{ 4\pi { \varepsilon }_{ 0 }r } =Q\times { 10 }^{ 11 }$

$\therefore 4\pi { \varepsilon }_{ 0 }r={ 10 }^{ -11 }...(i)$
Now,

$E=\cfrac { Q }{ 4\pi { \varepsilon }_{ 0 }{ r }^{ 2 } } =\cfrac { Q\times 4\pi { \varepsilon }_{ 0 } }{ (4\pi { \varepsilon }_{ 0 }{ r) }^{ 2 } } =\cfrac { Q\times 4\pi { \varepsilon }_{ 0 } }{ ({ 10 }^{ -11 })^{ 2 } } =4\pi { \varepsilon }_{ 0 }Q\times { 10 }^{ 22 }V{ m }^{ -1 }$ (Using (i))

The potential at a point due to a charge of

$5\times { 10 }^{ -7 }C$ located

$10cm$ away is:

0%
$3.5\times { 10 }^{ 5 }V$
0%
$3.5\times { 10 }^{ 4 }V$
0%
$4.5\times { 10 }^{ 4 }V$
0%
$4.5\times { 10 }^{ 5 }V$

Explanation

Here

$q=5\times { 10 }^{ -7 }C,r=10cm=0.1m\quad$
Potential,

$V=\cfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \cfrac { q }{ r } =\cfrac { 9\times { 10 }^{ 9 }\times 5\times { 10 }^{ -7 } }{ 0.1 } =4.5\times { 10 }^{ 4 }V$

The top of the atmosphere is about

$400kV$ with respect to the surface of earth, corresponding to an electric field that decreases with altitude. Near the surface of earth the field is about

$100V{m}^{-1}$ , but still don't get an electric shock, as we set out of our houses in to open because (assume the house is free from electric field)

0%
our body is a perfect insulator
0%
our body and ground form an equipotential surface
0%
the original euipotential surfaces of open air remain same
0%
none of these

A parallel plate capacitor is given a definite potential difference. Keeping the potential difference same, a slab of thickness 3 mm is placed between the plates. To do this, the distance between the plates is increased by 2.4 mm. Calculate the dielectric constant of the slab.

0%
10
0%
15
0%
5
0%
8

Explanation

$\dfrac{{\varepsilon A}}{d} = \dfrac{{{\varepsilon _0}A}}{{d' - t + \dfrac{r}{{{\varepsilon _r}}}}}$
or

$d = d' - t + \dfrac{t}{{{\varepsilon _r}}}$

$r - \dfrac{r}{{{\varepsilon _r}}} = d' - d = 2.4\,mm$

$r\left( {1 - \dfrac{1}{{{\varepsilon _r}}}} \right) = 2.4$

$1 - \dfrac{1}{{{\varepsilon _r}}} = \dfrac{{2.4}}{3}\, \Rightarrow \,{\varepsilon _r} = \dfrac{{30}}{6} = 5$

Consider a parallel plate capacitor having charge

$Q$ . Then

0%
There is no force acting on plates of the capacitor
0%
The electric field at outer side of plates is zero
0%
The electric field between the plates of the capacitor is $\dfrac{\sigma}{in_0}$
0%
The force between the plates is given by $\dfrac{Q^2}{2A \in_0}$

Explanation

In this case the charge will arrange in a way that it does not create any field outside as according to Gauss law. The net charge on the system containing both plates is zero. And in between the plates

$E=\cfrac{\sigma }{2 \varepsilon_{o}} + \cfrac{\sigma}{2 \varepsilon_{o}} = \cfrac{\sigma}{\varepsilon_{o}}$

Force between the plates

$= \cfrac{Q^{2}}{2A \varepsilon_{o}}$

A capacitor of capacitance

$C$ is connected to a cell of

$emf$ V and when fully charged, it is disconnected. Now the separation between the plates doubled. The change in flux of electric field through a closed surface enclosing the capacitor is :

0%
Zero
0%
$\dfrac{CV}{\varepsilon_0}$
0%
$\dfrac{CV}{2\varepsilon_0}$
0%
$\dfrac{2CV}{\varepsilon_0}$

Explanation

Flux

$= \dfrac{q_{in}}{\varepsilon_0}$
The two plates of the capacitor have equal and opposite charges. Hence, net charge enclosed by the given surface

$= 0$

$\therefore$ Flux is zero in both cases.
Hence change in flux

$= 0$ .

In a series combination of two capacitances

$C'$ and

$C(C> C)$ (as shown in the circuit)

0%
C' stores more energy than $C$
0%
$C$ stores more energy than $C'$
0%
potential difference across $C$ is more than that across $C'$
0%
potential difference across $C'$ is less than that across $C$

Explanation

In series combination of capacitor the potential across each capacitor is same and

$E= \cfrac{1}{2} CV^{2}$

$C >C'$ , so

$C$ stores more energy.

Two conducting plates A and B are placed parallel to each other. A is given

$Q_1$ and B a charge

$Q_2$ . Then

0%
The charge on the outer plate A is $\dfrac{Q_1 + Q_2}{2}$
0%
The charge on the outer plate B is $\dfrac{Q_1 + Q_2}{2}$
0%
The charge on inner plate A is $\dfrac{Q_1 - Q_2}{2}$
0%
The charge on inner plate B is $\dfrac{Q_1 - Q_2}{2}$

Work done in moving an object through an equipotential surface is

0%
Positive
0%
Negative
0%
Zero
0%
Depends on the field direction

A parallel plate capacitor with air as dielectric is charged to a potential 'V' using a battery. Removing the battery, the charged capacitor is then connected across an identical uncharged parallel plate capacitor filled with wax of dielectric constant 'K' the common potential of both the capacitor is

0%
V volts
0%
kV volts
0%
(k + 1) V volts
0%
$\dfrac{V}{{k + 1}}{\text{volts}}$

Explanation

Capacitance with air as dielectric

$C= \varepsilon_0 \dfrac{A}{d}$

For identical Capacitor Capacitance with wax as dielectric

$C'= \varepsilon \dfrac{A}{d}$

$\therefore C'=KC , \ where \ K=\dfrac{\varepsilon}{\varepsilon_0}$

When

$C \ and \ C'$ are connected in parallel charge is distributed and the potential difference is equal for the two capacitors

Resultant capacitance

$C_R=C+C'$ and the total charge

$Q=CV$

$\therefore$ Common \Potential \V=\dfrac{Q}{C_R}=\dfrac{CV}{C+C'}$$

$\therefore V=\dfrac{CV}{C+KC}=\dfrac{V}{K+1}$

An example of an equipotential surface in earth is

0%
A line passing through the centre of the earth connecting two points along the diameter
0%
A plane that passes through the circular section of the hemisphere of the earth
0%
A spherical surface at a distance of 1km from the surface of the earth with its centre at centre of earth
0%
A plane on the surface of the earth, which is a tangent to the earth

The electric field due to the electric potential V =

$(2x^{2}-4x)$ is

0%
$(4x+4)\hat{i}$
0%
$(4x-4)\hat{i}$
0%
$(-4x+4)\hat{i}$
0%
$(-4x-4)\hat{i}$

Explanation

$\vec { E } =-\dfrac{dV}{dx}$

Given,

$V={ 2x }^{ 2 }-4x$

$\therefore \quad \vec { E } =-\left( \dfrac { d }{ dx } ({ 2x }^{ 2 }-4x) \right) \hat i$

$=(- 4x+4 ) \hat i$

$\therefore$ Option (C) is the correct answer.

In the given figure each capacitor is equal to 45

$\mu F$ then the equivalent capacity between A and B in the given circuit is:

0%
$15 \mu F$
0%
$10 \mu F$
0%
$40 \mu F$
0%
$135 \mu F$

Explanation

ref. image I

$C = 45 \mu F$

On rearranging

ref. image II

$\Rightarrow C_{eq} = C + C + C$

$C_{eq} = 3C$

$\Rightarrow 135 \mu F$

Option (D)

1449470_1025375_ans_25fc876602ef4c4fb088a3c71860973a.png

Two insulated charged spheres of radii

${R}_{1}$ and

${R}_{2}$ having charges

${Q}_{1}$ and

${Q}_{2}$ respectively are connected to each other, then there is:

0%
no change in the energy of the sytem
0%
an increase in the energy of the system
0%
always a decrease in the energy of the system
0%
a decrease in energy of the system unless ${q}_{1}{R}_{2}={q}_{2}{R}_{1}$

Explanation

Two insulated charged spheres of radii

$R_1,R_2$ having charges

$q_1,q_2$ respectively are brought in contact with each other.

Charges will flow across the point of contact from where they are connected until their potential at their surfaces became same.

Energy is decreased because system changes to get stability.

Potential at their surfaces are same,
So,

$V_1=V_2$

$\dfrac{Kq_1}{R_1}=\dfrac{Kq_2}{R_2}$

$q_1R_2=q_2R_1$

Hence Proved.

Select the correct statements.

0%
The energy of a capacitor resides in the field between the plates
0%
The capacitance of a parallel plate capacitor does not depend on the metal of the plates
0%
If the current charging a capacitor is kept constant, the potential difference V across the capacitor varies with time according to the adjacent graph
0%
All of the above

Electrons are caused to fall through a potential difference of 1500 volt. If they were initially at rest, their final speed is

0%
$4.6 \times {10^7}m/s$
0%
$0.726 \times {10^7}m/s$
0%
$0.23 \times {10^2}m/s$
0%
$5.1 \times {10^9}m/s$

Explanation

V = 1500 V

u = 0, initial K.E = 0

Energy last by electron =

$V.e = 1500 \times 1.602 \times {10^{ - 19}}$

$\Rightarrow 2.403 \times {10^{ - 19}}\,\,joules$

$\begin{array}{l} mass\, \, of\, \, electrons\left( m \right) =9.1\times { 10^{ -31 } }kg \\ kinetic\, \, energy\, \, gained=P.E\, \, lost \\ \frac { 1 }{ 2 } m{ V^{ 2 } }=2.403\times { 10^{ -17 } } \\ { V^{ 2 } }=\frac { { 2\times 2.403\times { { 10 }^{ -17 } } } }{ { 9.1\times { { 10 }^{ -31 } } } } =\frac { { 4.806\times { { 10 }^{ +14 } } } }{ { 9.1 } } =0.528\times { 10^{ 14 } } \\ { V^{ 2 } }\sqrt { 0.528\times { { 10 }^{ 14 } } } =0.726\times { 10^{ 7 } }m/s \end{array}$

Three charges +4q,-q and +4q are kept on a straight line at position (0,0,0),(a,0,0) and (2a,0,0) respectively. Considering that they are free to move along the x-axis only

0%
All the charges are in stable equilibrium
0%
All the charges are in unstable equilibrium.
0%
Only the middle charge is in stable equilibrium.
0%
Only middle charge is in unstable equilibrium.

Explanation

Configuration of the charges : [Ref. image]

From symmetry and by using Columb's law

$\left ( F=\dfrac{kQ_{q}}{R^{2}} \right )$

one can show net force on each charge is zero.______(i)

also, potential energy of system,

$u=k\left [ \dfrac{(4q)(-q)}{a} +\dfrac{(4q)(-q)}{a}+\dfrac{(4q)(4q)}{2a}\right ]$

$u=0$ _____(ii)

(i) and (ii) imply ,

system is in stable equilibrium.

1048957_1019849_ans_632f737ee4c14f02a9c59d1eb3456913.png

The earth has volume 'V' and surface area 'A'; then its capacitance would be:

0%
$4\pi \in _0 \dfrac{A} {V}$
0%
$4\pi \in _0 \dfrac{V} {A}$
0%
$12\pi \in _0 \dfrac{V} {A}$
0%
$12\pi \in _0 \dfrac{A} {V}$

Explanation

Capacitance of an isolated spherical capacitor is,

$C=4 \pi\varepsilon _0R$ (

$R\ \rightarrow$ Radius of earth)

Now,

$A=4 \pi R^2$ and

$V=\dfrac{4}{3}\pi R^3$

$\Rightarrow R =\dfrac{3V}{A}$ .

which gives,

$C=12 \pi \varepsilon _0\dfrac{V}{A}$ .

CBSE Questions for Class 12 Medical Physics Electrostatic Potential And Capacitance Quiz 8 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Electrostatic Potential And Capacitance Quiz 8 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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