MCQ Questions for Class 12 Medical Physics Electrostatic Potential And Capacitance Quiz 1

The plates of a parallel-plate capacitor are not exactly parallel. The surface charge density, therefore,

0%
is smallest where the plates are closet
0%
is higher at the closer end
0%
will not be uniform
0%
each plate will have the same potential at each point

Explanation

If the plates are metallic, its a equipotential surface. So the potential will be same at each point on the plates.
Since

$Q = CV = A\epsilon V / d$
Hence, Q is inversely proportional to d. So at closer end, Q and Surface charge density would be highest.

In the given circuit, if potential difference between points

$A$ and

$B$ is

$4V$ , then, resistance

$X$ is:-

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$5 \Omega$
0%
$10 \Omega$
0%
$15 \Omega$
0%
$20 \Omega$

What is the S.I. unit of electric potential?

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ampere
0%
volt
0%
volt.m
0%
coulomb

A battery is used to charge a parallel plate capacitor till the potential difference between the plates becomes equal to the electromotive force of the battery. The ratio of the energy stored in the capacitor and the work done by the battery will be :

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$1$
0%
$2$
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$\dfrac{1}{4}$
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$\dfrac{1}{2}$

Explanation

energy in capacitor =

$\frac{1}{2} C V^2$
word done by battery =

$V Q = V CV = CV^2$
therefore their ratio is

$= \frac{1}{2}$

An electric charge

$10^{-3}\mu C$ is placed at the origin

$(0,0)$ of X - Y co-ordinate system. Two points A and B are situated at

$(\sqrt{2},\sqrt{2})$ and

$(2, 0)$ respectively. The potential difference between the points A and B will be:

0%
$9$ volt
0%
zero
0%
$2$ volt
0%
$4.5$ volt

Explanation

$V_A = \dfrac{1}{4 \pi \epsilon } \dfrac{ q}{r_A} = \dfrac{1}{4 \pi \epsilon } \dfrac{ q}{\sqrt{2+2}}= \dfrac{1}{4 \pi \epsilon } \dfrac{ q}{2}$

$V_B= \dfrac{1}{4 \pi \epsilon } \dfrac{ q}{r_B}= \dfrac{1}{4 \pi \epsilon } \dfrac{ q}{\sqrt{4+0}}= \dfrac{1}{4 \pi \epsilon } \dfrac{ q}{2}$

$V_A - V_B =0$

A parallel plat capacitor is made of two circular plates separated by a distance of

$5 mm$ and with a dielectric of dielectric constant

$2.2$ between them. When the electric field in the dielectric is

$3 \times 10^4$ V/m, the charge density of the positive plate will be close to:

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$3 \times 10^4 C/m^2$
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$6 \times 10^4 C/m^2$
0%
$6 \times 10^{-7} C/m^2$
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$3 \times 10^{-7} C/m^2$

Explanation

Answer is C.
By formula of electric field between the plates of a capacitor

$\displaystyle E = \dfrac{\sigma}{K\varepsilon_0}$

$\sigma = EK\varepsilon_0 = 3 \times 10^4 \times 2.2 \times 8.85 \times 10^{-12}$

$= 6.6 \times 8. 85 \times 10^{-8}$

$= 5.841 \times 10^{-7}$

$\cong 6 \times 10^{-7} C/m^2$

A sheet of aluminium foil of negligible thickness is introduced between the plates of a capacitor. The capacitance of the capacitor

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decreases
0%
remains unchanged
0%
becomes infinite
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increases.

Explanation

The capacitance without aluminium foil is

$C=\dfrac{A\epsilon_0}{d}$

The capacitance with aluminium foil is

$C'=\dfrac{A\epsilon_0}{d-t}$

If thickness

$(t)$ of foil is negligible so

$d-t \sim d.$ Hence,

$C=C'$

A parallel plate capacitor has a dielectric slab of dielectric constant K between its plates that covers 1/3 of the area of its plates, as shown in the figure. The total capacitance of the capacitor is C while that of the portion with dielectric in between is C

$_1$ . When the capacitor is charged, the plate area covered by the dielectric gets charge Q

$_1$ and the rest of the area gets charge Q

$_2$ . The electric field in the dielectric is E

$_1$ and that in the other portion is E

$_2$ . Choose the correct option/options, ignoring edge effects :

0%
$\displaystyle \frac{E_1}{E_2} = 1$
0%
$\displaystyle \frac{E_1}{E_2} = \frac{1}{K}$
0%
$\displaystyle \frac{Q_1}{Q_2} = \frac{3}{K}$
0%
$\displaystyle \frac{C}{C_1} = \frac{2+K}{K}$

Explanation

$E = V/d$

$E_1 / E_2=1$ (both parts have common potential difference)
Assume

$C_0$ be the capacitance without dielectric for whole capacitor.

$\displaystyle K \frac{C_0}{3} + \frac{2C_0}{3} = C$

$\displaystyle \frac{C}{C_1} = \frac{2+K}{K}$

$\displaystyle \frac{Q_1}{Q_2} = \frac{K}{2}$

What is the electric potential at a distance of 9 cm from 3 nC?

0%
$270 V$
0%
$3 V$
0%
$300 V$
0%
$30 V$

Explanation

Given :

$r = 9$ cm

$= 0.09$ m

Charge

$Q = 3$ nC

$= 3\times 10^{-9}$ C

Potential at a distance 9 cm,

$V = \dfrac{kQ}{r}$ where

$k = 9\times 10^9$

$\therefore$

$V = \dfrac{9\times 10^9 \times 3\times 10^{-9}}{0.09} = 300$ V

A parallel plate air capacitor of capacitance C is connected to a cell of emf V and then disconnected from it. A dielectric slab of dielectric constant K, which can just fill the air gap of the capacitor, is now inserted in it. Which of the following is incorrect?

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The change in energy stored is $\frac {1}{2}CV^2\left (\frac {1}{K}-1\right )$
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The charge on the capacitor is not conserved
0%
The potential difference between the plates decreases K times
0%
The energy stored in the capacitor decreases K times

Explanation

$C_f = KC_i$

$Q_i =C_iV$

$U_i = \frac{1}{2}C_iV^2$

$U_f = \frac{1}{2}\frac{Q_f ^2}{C_f}$
After the capacitor is disconnected from the battery, charge on the capacitor cannot change.

$\therefore Q_f =Q_i$

$\therefore U_f = \frac{1}{2} \frac{Q_f ^2}{C_f}= \frac{1}{2}\frac{C_i ^2V^2}{KC_i}$

$\Delta U = U_f -U_i = \frac{1}{2}\frac{C_i ^2V^2}{KC_i} - \frac{1}{2}C_i V^2= \frac{1}{2}C_i V^2 ( \frac{1}{K} -1)=\frac{1}{2}C V^2 ( \frac{1}{K} -1)$ since

$C_i= C$ as given

Option(A) is correct.

Option (B) :
Charge on the capacitor remains same since battery is disconnected.

Option (C):
C increases K times.
Since

$Q = CV$ and Q remains same, V decreases K times.

Option D:
The energy stored in the capacitor

$U_f$ decreases K times.

The diagrams below show regions of equipotentials.
A positive charge is moved from A to B in each diagram.

0%
Maximum work is required to move $q$ in figure (b)
0%
Maximum work is required to move $q$ in figure(c)
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In all the cases the work done is the same.
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Minimum work is required to move $q$ in figure(a)

Two parallel metal plates having charges +Q and -Q face each other at a certain distance between them. If the plates are now dipped in kerosene oil tank, the electric field between the plates will :

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become zero
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increase
0%
decrease
0%
remains same

Explanation

Electric field between two parallel plates placed in vacuum is given by
E =

$\displaystyle \frac{\sigma}{\varepsilon_0}$
In a medium of dielectric constant K, E' =

$\displaystyle \frac{\sigma}{\varepsilon_0 K}$ For kerosene oil K > 1

$\Rightarrow$ E' is less than E

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
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Both Assertion and Reason are incorrect

In figure, a particle has mass

$m=5g$ and charge

$q'=2\times {10}^{-9}C$ starts from rest at point

$a$ and moves in a straight line to point

$b$ . What is its speed

$v$ at point

$b$ ?

0%
$2.65\ {cms}^{-1}$
0%
$3.65\ {cms}^{-1}$
0%
$4.65\ {cms}^{-1}$
0%
$5.65\ {cms}^{-1}$

Explanation

Potential at a point

$p$ due to the charge

$q$ ,

$V_p = \dfrac{Kq}{r}$

Where,

$K = \dfrac{1}{4\pi \epsilon_o} =9\times 10^9$

Total potential at point

$a$ ,

$V_a = \dfrac{9 \times 10^9 \times 3 \times 10^{-9}}{10^{-2}} + \dfrac{9\times 10^9 \times (-3 \times 10^{-9})}{2 \times 10^{-2}} = 13.5 \times 10^2$ V

Total potential at point

$b$ ,

$V_b = \dfrac{9 \times 10^9 \times 3 \times 10^{-9}}{2\times 10^{-2}} + \dfrac{9\times 10^9 \times (-3 \times 10^{-9})}{ 10^{-2}} = -13.5 \times 10^2$ V

Potential difference between point a and b,

$V_{ab} = V_a - V_b = 27 \times 10^2$ V

Work done in moving the charge

$q'$ from a to b,

$W_{ab} = q' V_{ab}$

$W_{ab} = 2\times 10^{-9} \times 2700 = 5.4 \times 10^{-6}$ J

From work-energy theorem :

$W_{ab} = \Delta K.E$

$\Rightarrow$

$5.4 \times 10^{-6} = \dfrac{1}{2} \times 0.005 \times v^2$

$\Rightarrow v = 0.0465$

$ms^{-1} = 4.65$

$ms^{-1}$

Two identical conductors of copper and aluminium are placed in an identical electric field. What is the magnetic of induced charge in the aluminium?

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Less than in copper
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Equal to that in copper
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Greater than in copper
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Zero

An electric charge

${10}^{-3}\mu C$ is placed at the origin

$(0,0)$ of

$(x,y)$ co-ordinate system. Two points

$A$ and

$B$ are situated at

$(\sqrt {2}, \sqrt {2})$ and

$(2,0)$ respectively. The potential difference between the points

$A$ and

$B$ will be:

0%
$4.5\ V$
0%
$9\ V$
0%
$zero$
0%
$2\ V$

Explanation

Given :

$q = 10^{-3} \mu C = 10^{-9} C$

Distance of A from origin

$r_A = \sqrt{(\sqrt{2} - 0)^2 + (\sqrt{2} - 0)^2} = 2$

Distance of B from origin

$r_B = 2$

Thus potential due to charge at point A

$V_A = \dfrac{Kq}{r_A}$ where

$K = \dfrac{1}{4\pi \epsilon_o} = 9 \times 10^9$

$\therefore$

$V_A = \dfrac{9 \times 10^9 \times 10^{-9}}{2} = 4.5$ V

Potential due to the charge at B

$V_B = \dfrac{Kq}{r_B} = \dfrac{9\times 10^9 \times 10^{-9}}{2} = 4.5$ V

Potential difference between A and B

$V_{AB} = V_A- V_B = 0$

A parallel plate capacitor with air as a dielectric has capacitance

$C$ . A slab of dielectric constant

$K$ , having same thickness as the separation between the plates is introduced so as to fill one-fourth of the capacitor as shown in the figure. The new capacitance will be:

0%
$(K + 3)\dfrac {C}{4}$
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$(K + 2)\dfrac {C}{4}$
0%
$(K + 1)\dfrac {C}{4}$
0%
$\dfrac {KC}{4}$

Explanation

The two condensers with

$K$ and with air are in parallel. With air,

$C_{1} = \dfrac {\epsilon_{0}}{d}\left (\dfrac {3A}{4}\right ) = \dfrac {3\epsilon_{0}A}{4d}$
With medium,

$C_{2} = \dfrac {\epsilon_{0}K}{d}\left (\dfrac {A}{4}\right ) = \dfrac {\epsilon_{0}AK}{4d}$

$\therefore C' = C_{1} + C_{2}$
or

$C' = \dfrac {3\epsilon_{0}A}{4d} + \dfrac {\epsilon_{0}AK}{4d} = \dfrac {\epsilon_{0}A}{d}\left [\dfrac {3}{4} +\dfrac {K}{4}\right ]$
or

$C' = \dfrac {C}{4} (K + 3) \left [\because C = \dfrac {A\epsilon_{0}}{d}\right ]$ .

$100J$ of work is done when

$2 \mu C$ charge is moved in an electric field between two points. The p.d. between the points is

0%
$2\times10^{-4}V$
0%
$2\times10^{-8}V$
0%
$2\times10^{-6}V$
0%
$5\times10^{7}V$

Explanation

We know work done

$=q\Delta V$
Where

$\Delta V$ is change in potential

$V_2-V_1$

$100=\Delta V\times 2\times 10^{-6}$

$50\times 10^6=\Delta V$

$\therefore p\cdot d= 5\times 10^7 V$

In the given figure each plate of capacitance C has value of charge -

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$CE$
0%
$\cfrac { { CER }_{ 1 } }{ R_{ 2 }-r }$
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$\cfrac { { CER }_{ 2 } }{ R_{ 2 }+r }$
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$\cfrac { { CER }_{ 1 } }{ R_{ 1 }-r }$

In a charged capacitor the energy is stored in:

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both in positive and negative charges
0%
positive charges
0%
the edges of the capacitor plates
0%
the electric field between the plates

Explanation

We know energy per unit volume stored in electric field

$=\frac {1}{2} \varepsilon_0 E^ 2$
in parallel plate capacitor

$E=\frac {6}{\varepsilon_0}$
Energy density

$=\frac {1}{2}\varepsilon_0 \frac {6^2}{\varepsilon_o^2}$

$Energy = \frac {1}{2}\varepsilon_0 \frac {1}{\varepsilon_0}(\frac {Q}{A})^2\cdot (A\cdot d)\rightarrow volume$

$E=\frac {1}{2}\frac {1}{\varepsilon_0}\frac {Q^2d}{A}$
We know

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

$E=\frac {1}{2}\frac {Q^2}{C} (Q=CV)$

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

When air is replaced by a dielectric medium of constant $K$ , the capacity of the condenser:

0%
increases $K$ times
0%
increases $K^{2}$ times
0%
remains unchanged
0%
decreases $K$ times

Explanation

We know, capacity:

$C=\dfrac {\varepsilon_0 A}{d}$
Now when air is replaced by dielectric medium of constant

$k$

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

This implies that the capacitance increases

$K$ times.

Space between the plates of a parallel plate capacitor is filled with a dielectric slab. The capacitor is charged and then the supply is disconnected to it. If the slab is now taken out , then:

0%
work is not done to take out the slab
0%
energy stored in the capacitor reduces
0%
potential difference across the capacitor is decreased
0%
potential difference across the capacitor is increased

Explanation

When a capacitor is charged and then the supply is disconnected charge present on it becomes constant (law of conservation of charge)
Initial capacitance

$C=\frac {k\varepsilon_0 A}{d}$ (with slab)
Final capacitance

$C=\frac {\varepsilon_0 A}{d}$ (slab removed)
We know

$Q= CV$
As C decreases and Q remains constant V has to increase.

A parallel plate condenser is charged by connecting it to a battery. The battery is disconnected and a glass slab is introduced between the plates. Then,

0%
potential increases
0%
electric intensity incrases
0%
energy decreases
0%
capacity decreases

Explanation

$\rightarrow$ When the capacitor is charged and battery is removed the charge present on capacitor remains constant. (Law of conservation of charge)

$\rightarrow$ We know

$C=\dfrac {\varepsilon_0 A}{d}$
When slab is introduced,

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

$C$ increases and

$Q$ is constant.
We know:

$Q=CV$

$\therefore$

$V$ has to decrease.
We know energy store in capacitor

$=\dfrac{1}{2} QV$

$\therefore$ as V decreases, energy decreases.

Fill in the blank:

In order to increase the capacity of a parallel plate condenser, one should introduce a sheet of

$\underline{\hspace{0.5in}}$ between the plates (assume that the space is completely filled).

0%
Mica
0%
Tin
0%
Copper
0%
Stainless Steel

The capacity of a parallel plate air condenser is

$2\ \mu F$ . If a dielectric of dielectric constant

$4$ is introduced between the plates, its new capacity is:

0%
$1.5\ \mu F$
0%
$0.5\ \mu F$
0%
$8\ \mu F$
0%
$6\ \mu F$

Explanation

We know

$C=\dfrac{\epsilon_{0}A}{d}$
When dielectric is added:

$C'=\dfrac{K \epsilon_{0}A}{d}$

$=4\times 2\ \mu F$

$=8\mu F$

A metal plate of thickness half the separation between the capacitor plates of capacitance C is inserted. The new capacitance is:

0%
C
0%
$\frac{C}{2}$
0%
zero
0%
2C

Explanation

Initial capacitance

$=\frac {\varepsilon_0 A}{d}$
Now after metal plate is inserted it behaves as two capacitor plates in series with capacitance

$\frac {\varepsilon_0 A}{(\frac {d}{a})}$
We know when capacitance are in series

$\frac {1}{C_{eff}}=\frac {1}{C_1}+\frac {1}{C_2}$

$\Rightarrow \frac {1}{C_{eff}}=\frac {d}{4\varepsilon_0 A}+\frac {d}{4\varepsilon_0 A}$

$\Rightarrow C_{eff}=\frac {2\varepsilon_0 A}{d.}$

$\therefore Now \ C_{eff}= 2C$

In an electrical circuit, which of the following quantities is analogous to temperature?

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Potential
0%
Resistance
0%
Current
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Charge

A parallel plate capacitor is first charged and then isolated , and a dielectric slab is introduced between the plates. The quantity that remains unchanged is:

0%
charge $Q$
0%
potential $V$
0%
capacity $C$
0%
energy $U$

Explanation

When the capacitor is kept at a voltage, it gains charge.
Now when the system is isolated, the charge present on capacitor cannot change because of law of conservation of charge.

$\therefore$ Charge always remains constant in isolated systems.

Two capacitors of capacity

$C_1$ and

$C_2$ are connected in parallel, then the equivalent capacity is:

0%
$C_1+C_2$
0%
$C_1C_2/(C_1+C_2)$
0%
$C_1/C_2$
0%
$C_2/C_1$

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)

$C=C_1+C_2$

$1$ volt is equal to

0%
${1 \ Joule}$
0%
$\dfrac{1 \ Joule}{1 \ Coulomb}$
0%
$\dfrac{1 \ Joule}{1 \ meter}$
0%
$\dfrac{1 \ Newton}{1 \ Coulomb}$

State whether given statement is True or False.
Electric potential at a point in an electric field is defined as the work done in moving a unit positive charge from infinity to that point.

0%
True
0%
False

What is the shape of the equipotential surface for the line charge ?

0%
Sphere
0%
Cylinder
0%
It will depend upon the type of charge
0%
Can't say

The electric volt is a measure of electrical potential. Identify which of the following can be defined as a volt.

0%
Opposition to electrical motion
0%
Number of particles in motion
0%
Work per unit charge
0%
Field strength per unit of force
0%
Electrostatic discharge

Explanation

eV is defined as the kinetic energy gained by a unit charge

$e$ when it is accelerated by a potential difference of 1 volt. As we know kinetic energy is the measure of work done, therefore, volt can be defined as work done per unit charge.

A one microfarad capacitor of a TV is subjected to 4000 V potential difference. The energy stored in capacitor is :

0%
8J
0%
16 J
0%
$4\times 10^{-3}J$
0%
$2\times 10^{-3}J$

Explanation

Given, capacitor

$C=10^{-6} F$ and potential difference

$V=4000 V$
The energy stored in capacitor

$=\dfrac{1}{2}CV^2=\dfrac{1}{2}\times 10^{-6}\times (4000)^2=8 J$

The equivalent capacity between the points ‘A’ and ‘B’ in the following figure will be:

0%
$9 \mu F$
0%
$1 \mu F$
0%
$4.5 \mu F$
0%
$6 \mu F$

Explanation

Hint:

If two capacitors with capacitances ${{C}_{1}}\,and\,{{C}_{2}}$ are connected in parallel then there equivalent capacitance is given as,

${{C}_{eq.}}={{C}_{1}}\,+\,{{C}_{2}}$

Step 1: Simplify the Circuit.

The given circuit can be simplified as shown in figure with three capacitors ( ${{C}_{1}}={{C}_{2}}={{C}_{3}}=3\mu F$ ) in parallel between A and B.

Step 2: Calculate equivalent capacitance between A and B.

Equivalent capacitance between A and B is,

${{C}_{eq.}}={{C}_{1}}+{{C}_{2}}+{{C}_{3}}$

$\Rightarrow {{C}_{eq.}}=(3+3+3)\mu F$

$\Rightarrow {{C}_{eq.}}=9\mu F$

Thus, equivalent capacitance between A and B is $9\mu F$ .

The work done in moving a single positive charge from infinity to a point is called ..... at that point

0%
Electricity
0%
Electric potential
0%
Potential gradient
0%
None of these

Explanation

The work done is moving a single positive charge from infinity to a point is called electric potential at that point.

Work done

$=V=E.d$

To bring a unit positive charge from infinity to a point in an electric field, some work has to done, which is called:

0%
potential energy
0%
electric potential
0%
electric field
0%
electric induction

Three capacitors of $3 \mu F, 2 \mu F$ and $6 \mu F$ are connected in series. When a battery of 10V is connected to this combination then charge on $3 \mu F$ capacitor will be :

0%
$5 \mu C$
0%
$10 \mu C$
0%
$15 \mu C$
0%
$20 \mu C$

Explanation

$q=CV\ \ \ \dfrac{1}{C_{eff}}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}$

$\dfrac{1}{C_{ef}}=\dfrac{1}{3}+\dfrac{1}{2}+\dfrac{1}{6}$

$\therefore C_{ef}=1 \mu F$

$\therefore q=C_{ef} V = 10 \mu C$
q is same for all capacitors

$\therefore$ q on any capacitor is

$10 \mu C$

For capacitors in the series combination, the total capacitance C is given by

0%
$C=(\cfrac{1}{C_1}+\cfrac{1}{C_2} + ......)$
0%
$C = C_{1} + C_{2} +$ .......
0%
$\cfrac{1}{C}=(\cfrac{1}{C_1}+\cfrac{1}{C_2}+.....)$
0%
$\cfrac{1}{C} = C_{1} + C_{2} +$ ........

There is no flow of current between two equally charged identical spheres when connected, because, they have the same

0%
quantity of charge
0%
potential
0%
capacity
0%
ratio of potential per unit charge

Name the unit of electrical potential :

0%
Coulomb
0%
Watt
0%
Joule
0%
Volt

Electric potential is :

0%
a scalar quantity
0%
a vector quantity
0%
neither scalar nor vector
0%
sometimes scalar and sometimes vector

Explanation

Electric potential is the electric potential energy per unit charge. In equation form,

$V=U/q,$ where

$U$ is the potential energy,

$q$ is the charge, and

$V$ is the electric potential. Since both the potential energy and charge are scalar quantities, so does the potential.

1 volt = ?

0%
1 joule
0%
1 joule per coulomb
0%
1 coulomb per metre
0%
1 newton per coulomb

Explanation

Using:

$V = \dfrac{W}{q}$

$\implies$ 1 volt is equal to 1 joule per coulomb

When a dielectric is introduced between the plates of a condenser, the capacity of condenser :

0%
increases
0%
decreases
0%
remains same
0%
none of these

Explanation

If the empty Condensor has capacity

$C$ , then its capacity with dielectric is given by

$C' = kC$ , where

$k$ is the dielectric constant of the dielectric material.

$k$ can never be less than

$1$ .

Which material sheet should be placed between the plates of a parallel plate condenser in order to increase its capacitance ?

0%
mica
0%
copper
0%
tin
0%
iron

Eight identical spherical mercury drops charged to a potential of 20 v each are coalesced into a single spherical drop:

0%
The internal Energy of the system remains the same
0%
The new potential of the drop is 80 V
0%
Internal Energy of he system decreases
0%
The potential remains the same i.e. 20 v

Explanation

$V=\dfrac {kq}{r}=20V$

$\dfrac {4}{3}\pi R^3=8\times \dfrac {4}{3}\pi r^3\Rightarrow R=2r$

$Q'=8q$

$V'=\dfrac {kQ'}{R}=\dfrac {k\times 8q}{2r}=\dfrac {4kq}{r}=4\times 20$

$V'=80 V$

When a thin mica sheet is placed between the plates of a condenser then the amount of charge, so compared to its previous value, on its plates will become:

0%
unchanged
0%
zero
0%
less
0%
more

One Volt is equal to:

0%
1 Joule
0%
1 Newton/Coulomb
0%
1 Joule/Coulomb
0%
1 Coulomb/Newton

Explanation

The volt is a measure of electric potential. One volt is defined as the difference in electric potential between two points of a conducting wire when an electric current of one ampere dissipates one watt of power between those points. It is also equal to the potential difference between two parallel, infinite planes spaced 1 meter apart that create an electric field of 1 newton per coulomb. Additionally, it is the potential difference between two points that will impart one joule of energy per coulomb of charge that passes through it. It can be expressed in terms of SI base units (m, kg, s, and A)

$1 volt = \dfrac{1 Joule}{Coulomb}$

A number of capacitors, each of equal capacitance

$C$ , are arranged as shown in the figure. The equivalent capacitance between

$A$ and

$B$ is :

0%
${ n }^{ 2 }C$
0%
$(2n+1)C$
0%
$\cfrac { (n-1)n }{ 2 } C\quad$
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$\cfrac { (n+1)n }{ 2 } C\quad$

Explanation

There are n groups

$C_1=C$

$C_2=2C$

$C_3=3C$
Likewise

$C_n=nC$

$C_{eq}=C_1+C_2+C_3+......+C_n$

$C_{eq}=C+2C+3C+............nC=\frac{n(n+1)C}{2}$

The capacity of a parallel plate condenser is inversely proportional to ___________.

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Area of each plate
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Dielectric constant
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Permittivity of medium
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Distance between two plates

Explanation

Capacitance is the ability of a component or circuit to collect and store energy in the form of an electrical charge. It depends upon the physical dimensions of the capacitor.

For a parallel plate capacitor, it is given as

$C=\dfrac{A\epsilon_0}{d}$

$\implies C\propto \dfrac{1}{d}$

CBSE Questions for Class 12 Medical Physics Electrostatic Potential And Capacitance Quiz 1 - 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 1 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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