MCQ Questions for Class 12 Medical Physics Electrostatic Potential And Capacitance Quiz 11

A parallel place condenser has conducting of rasines 12

$\mathrm { cm }$ seperated by a distance of 5

$\mathrm { mm }$ . It is charged with a constant changing curreat of 0.16

$\mathrm { A }$ the rate at which the potential difference between the plates change is

0%
$1 \times 10 ^ { \circ } \mathrm { Vs } ^ { - 1 }$
0%
$2 \times 10 ^ { 10 } \mathrm { Vs } ^ { - 1 }$
0%
$3 \times 10 ^ { 12 } \mathrm { Vs } ^ { - 1 }$
0%
$2 \times 10 ^ { 9 } \mathrm { Vs } ^ { - 1 }$

Consider a uniformly charged sheet

$ABCD$ , which is a part of an equilateral triangular sheet of a side a as shown in the figure. Choose the correct options regarding the electric field

$E$ at point

$O$ due to this sheet.

0%
Magnitude of $\vec {E}$ , increases with the increase in a (Keeping charge density same)
0%
Magnitude of $E$ , decreases with increase in a (keeping total charge same)
0%
If charge density is $\sigma$ and $a = 1\ m$ , magnitude of $E$ is equal to $\dfrac {7\sigma}{44\epsilon_{0}} ln \sqrt {2}$
0%
If charge density is $\sigma$ and $a = 2\ m$ , magnitude of $E$ is equal to $\dfrac {7\sigma}{22\epsilon_{0}} ln \sqrt {2}$

If a thin metal foil of the same area is placed between the two plates of a parallel plate capacitor capacitance C, then new capacitance will be

0%
C
0%
2C
0%
3C
0%
4C

Explanation

We have,

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

After placement of plate of equal area in between the two plates, we have two capacitors in series.

Value of each capacitor,

$C_1=\dfrac{\epsilon_0 A}{d/2}$

$C_1=\dfrac{2\epsilon_0 A}{d}=2C$

Equivalent capacitance,

$C_{eq}=\dfrac{2C \times 2C}{2C+2C}$

$\implies C_{eq}=C$

Four metal conductors having different shapes are mounted on insulating stands and charged.The one which is best suited to retain the charges for a longer time is.

0%
A sphere
0%
Cylindrical
0%
Pear
0%
Lightning conductor.

The electric potential at the apex p of regular pyramid of length '

$l$ ' due to system of charges q, -2q, and 3q is

0%
$\dfrac{2q}{\pi \epsilon_0 l}$
0%
$\dfrac{q}{4 \pi \epsilon_0 l}$
0%
$\dfrac{q}{2 \pi \epsilon_0 l}$
0%
$\dfrac{q}{\pi \epsilon_0 l}$

two similar capacitor are connected to potential v in parallel order by separating them and joining them in series

0%
the potential on fees plated will be doubled
0%
the charge on the face free plates will increase
0%
the plates in contact will lose their charge
0%
more energy will be stored in the system.

Five identical metal plates 1, 2, 3, 4 and 5 each of area A on one side are fixed parallel and equidistant (d) to each other. The plates 1 and 4 are joined by a conductor, and plates 3 and 5 are also joined by a conducter as shown in figure. Then, the capacitance of this system between A and B is-

0%
$\dfrac{5\in_0A}{d}$
0%
$\dfrac{4\in_0A}{d}$
0%
$\dfrac{5\in_0A}{3d}$
0%
None of these

If the area of one of the plates of a parallel plate capacitor is half that of the other, then charges on its plates are:

0%
$+Q$ and $Q$
0%
$+Q$ and $+Q$
0%
$4Q$ and $Q/2$
0%
$+ Q/2$ and $Q$

Two capacitor of capacity

$C_{1}$ and

$C_{2}$ are connected in series. The combined capacity

$C$ is given by

0%
$C_{1} + C_{2}$
0%
$C_{1} - C_{2}$
0%
$\dfrac {C_{1}C_{2}}{C_{1} + C_{2}}$
0%
$\dfrac {C_{1} + C_{2}}{C_{1}C_{2}}$

A parallel plate capacitor has

$1 \mu F$ capacitance. One of its two plates is given

$+2 \mu C$ charge and the other plate,

$+ 4 \mu C$ charge. The potential difference developed across the capacitor is :-

0%
$5V$
0%
$2V$
0%
$3 V$
0%
$1 V$

Explanation

$\because$ Electric fd. due to cap plate is

$\dfrac{\sigma}{2\varepsilon _0}$

Since both the plate has given positive charges.

$\boxed{E=\dfrac{\sigma}{2\varepsilon_0}}$

Net Electric .field.

$\dfrac{\sigma_2}{2\varepsilon_0}-\dfrac{\sigma_1}{2\varepsilon_0}$

$\boxed{\sigma=\dfrac{Q}{A}}$

$E_{net}=\dfrac{Q_2}{2AE_0}-\dfrac{Q_1}{2AE_0}$

$E_{net}=\dfrac{1}{1AE_0}(Q_2-Q_1)$ here

$Q_2=4\mu c$ Here

$Q_2=2 \mu c$

$\boxed{E_{net}=\dfrac{1}{2AE_0}(4-2)=\dfrac{1}{AE_0}}$

Also

$V=ED$

$=E_{net}\times d=\dfrac{1}{AE_0}\times d=\dfrac{1}{c}$

$\dfrac{1}{1}$

$\because C=\dfrac{E_0A}{d}$

$\because C=1\mu F$

$\boxed {\therefore V=1V}$

$V = \dfrac{Q}{C} = \dfrac{1\mu C}{1\mu F} = 1$ Volt

1246136_1613996_ans_890e21936bfd4761b6a307e423519947.PNG

The equivalent capacity of the combination shown in figure is?

0%
C
0%
$2C$
0%
$3/2C$
0%
$C/2$

Two identical drops of a liquid are charged to the same potential of

$100$ V. They are then merged into one large drop, the potential of the large drop is?

0%
$172$ V
0%
$193$ V
0%
$159$ V
0%
$100$ V

A neutral spherical copper particle has a radius of

$10\ nm (1nm = 10^{-9}m)$ . It gets charged by applying the voltage slowly adding one electron at a time. Then the graph of the total charge on the particle vs the applied voltage would look like:

Explanation

For spherical surface

Potnetial

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

$\because$ charge increases in discrete manner, so till the time charge is same on sphere, the potential will remain same.

1433491_1688481_ans_812f6d111b464b7da8b426d6cb09dd56.png

When a bird sits on a very high voltage cable.

0%
Its feathers ten to spead
0%
Its feathers tend to compress
0%
It receives an electric shock
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Neither its feathers have any effect nor electric shock is received by it

An external agent pulls a unit positive charge from infinity to a point, then the potential of that point is

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Positive
0%
Negative
0%
May be positive or may be negative
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Zero

A semicircular wire of radius

$a$ having

$\lambda$ as charge per unit length is shown in the figure. The electric potential at the centre of the semicircular wire is

0%
$\lambda/\epsilon_{0}$
0%
$\lambda/4\pi \epsilon_{0}R$
0%
$\lambda/4\epsilon_{0}$
0%
None of these

A dielectric slab is inserted between the plates of a capacitor. The charge on the capacitor is

$Q$ and the magnitude of the induced charge on each surface of the dielectric is

$Q'$ .

0%
$Q'$ must be smaller than $Q$
0%
$Q'$ must be larger than $Q$
0%
$Q'$ must be equal to $Q$ .
0%
None

Explanation

The relation between the induced charge and the charge of the capacitor is given as:

$Q'=Q\left(1-\dfrac1K\right)$

The dielectric constant is always greater than one. So, the value of the induced cvharge will always be less than Q.

Option

$D$ is correct.

In a region electric field is parallel to x-axis. The equation of equipotential surface is

0%
$y = C$
0%
$x = C$
0%
$z = C$
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None of these

Two condensers

$C_1$ and

$C_2$ in a circuit are joined as shown in Fig. The potential of point A is

$V_1$ and that of B is

$V_2$ . The potential of point D will be?

0%
$\dfrac{1}{2}(V_1+V_2)$
0%
$\dfrac{C_1V_2+C_2V_1}{C_1+C_2}$
0%
$\dfrac{C_1V_1+C_2V_2}{C_1+C_2}$
0%
$\dfrac{C_2V_1-C_1V_2}{C_1+C_2}$

The effective capacitance between points X and Y in Fig., assuming

$C_2=10\mu F$ and that outer capacitors are all

$4\mu F$ each, is?

0%
$1\mu F$
0%
$3\mu F$
0%
$4\mu F$
0%
$5\mu F$

The equivalent capacitance between

$4$ and

$5$ is?

0%
$\left(\dfrac{3C}{4}\right)$
0%
$\left(\dfrac{3C}{2}\right)$
0%
$\left(\dfrac{3C}{5}\right)$
0%
$\left(\dfrac{5C}{4}\right)$

Two conducting sphere of radii

$r_1$ and

$r_2$ have same electric field near their surfaces. The ratio of their electrical potentials is

0%
$r_1^2/r^2_2$
0%
$r_2^2/r^1_2$
0%
$r_1/r_2$
0%
$r_2/r_1$

Explanation

$E_1=E_2 \Rightarrow \frac{kQ_1}{r_1^2}= \frac{kQ_2}{r_2^2}$

$\frac { { Q }_{ 1 } }{ { Q }_{ 2 } } =\left( \frac { { r }_{ 1 } }{ { r }_{ 2 } } \right) ^{ 2 }$

$\frac { { V }_{ 1 } }{ { V }_{ 2 } } =\frac { Q_{ 1 } }{ r_{ 1 } } \times \frac { { r }_{ 2 } }{ { Q }_{ 2 } } =\left( \frac { { r }_{ 1 } }{ { r }_{ 2 } } \right) ^{ 2 }\frac { { r }_{ 2 } }{ { r }_{ 1 } } =\frac { { r }_{ 1 } }{ { r }_{ 2 } }$

If the switch

$S_2$ is also closed. Match the table.

Capacitor	Charge
(m) $C_1$	(q) $\left(\dfrac{84}{5}\right)\mu C$
(n) $C_2$	(r) $\left(\dfrac{72}{5}\right)\mu C$
(o) $C_3$	(s) $\left(\dfrac{54}{5}\right)\mu C$
(p) $C_4$	(t) $\left(\dfrac{42}{5}\right)\mu C$

0%
(m, q) (n, r) (o, s) (p, t)
0%
(m, t) (n, s) (o, r) (p, q)
0%
(m, t) (n, q) (o, s) (p, r)
0%
(m, r) (n, s) (o, t) (p, q)

The capacitance between

$1$ and

$3$ is?

0%
$\left(\dfrac{3C}{4}\right)$
0%
$\left(\dfrac{3C}{2}\right)$
0%
$\left(\dfrac{5C}{2}\right)$
0%
$\left(\dfrac{5C}{4}\right)$

The equivalent capacitance of the network between points a and b is?

0%
$2\mu F$
0%
$4\mu F$
0%
$6\mu F$
0%
$8\mu F$

The potential at point B is

0%
-738 V
0%
-323 V
0%
-705 V
0%
-120 V

Explanation

Electric Potential:
At

$B : V_B = k ( \dfrac {q_1}{r_1} +\dfrac {q_2}{r_2})$

$= k(\dfrac {2.40 \times 10^{-9} C}{0.08m} + \dfrac { -6.50 \times 10^{-9} C}{0.06 m} ) = -705 V$

In the circuit shown in the figure, the switch can be shifted to positions

$'1'$ and

$'2'$ . The charge on capacitor

$C_1$ when the switch is at position

$'1'$ is?

0%
$120\mu C$
0%
$240\mu C$
0%
$360\mu C$
0%
$80\mu C$

The charge appearing on

$C_2$ is?

0%
$E\left(\dfrac{C_3C_4}{C_1+C_2}\right)$
0%
$E\left(\dfrac{C_1C_2}{C_1+C_2}\right)$
0%
$E\left(\dfrac{C_1C_2}{C_3+C_4}\right)$
0%
$E\left(\dfrac{C_3C_4}{C_3+C_4}\right)$

With

$420$ V across a and b, the value of

$(V_c-V_d)$ is?

0%
$24.6$ V
0%
$46.7$ V
0%
$18$ V
0%
$72$ V

The potential at point A is

0%
-738 V
0%
-323 V
0%
-705 V
0%
-120 V

Explanation

Electric Potential:

$A : V_A = k ( \dfrac {q_1}{r_1} + \dfrac {q_2}{r_2})= k( \dfrac { 2.40 \times 10^{-9} C }{ 0.05 m } + \dfrac { -6.50 \times 10^{-9} C} {0.05 m } ) = -738 V$

Find the charge on

$C_1$ when only

$S_1$ is closed.

0%
$16\mu C$
0%
$9\mu C$
0%
$6\mu C$
0%
$8\mu C$

The electric field in a certain region is

$A/x^3$ . Then, the potential at a point (x,y,z), assuming the potential at infinity to be zero, is

0%
zero
0%
$A/2x^2$
0%
$3 A/x^4$
0%
$A/x^2$

An unchanged conductor A is brought near a positively charged conductor B. Then

0%
The charge on B will increase but the potential of B will not change
0%
the charge on B will not change but the potential of B will decrease
0%
the charge on B will decrease but the potential of B will not change
0%
the charge on B will not change but the potential of B will increase

The graph

$E_x$ versus x will be

The potential at the common center is

0%
$\frac { \sqrt {2} }{ \pi \epsilon_0} \frac {Q( R+ r)}{(R^2 + r^2 ) }$
0%
$\frac {1}{ 2 \pi \epsilon_0} \frac {Q ( R +r) }{ ( R^2 +r^2 )}$
0%
$\frac {1}{ 4 \pi \epsilon_0} \frac {Q( R +r) }{ (R^2 +r^2 )}$
0%
$\frac {1}{ \pi \epsilon_0} \frac {Q (R -r)}{(R^2 +r^2) }$

Explanation

$q_1$ and

$q_2$ are the charges on the spheres of radii r and R respectively , then, by conservation of charge

$q_1 +q_2 =Q ..................(i)$
But

$\sigma_1 = \sigma_2$

$\frac {q_1}{ 4 \pi r^2} = \frac {q_2}{ 4 \pi R^2 }$

$\frac {q_1}{q_2} = \frac {r^2}{R^2} .............(ii)$
From Eqs (I) and (ii) , we get

$q_1 = \frac {Qr^2}{ r^2 +R^2}$ and

$q_2 = \frac {QR^2}{r^2+ R^2}$
hence , potential at common center,

$V = V_1 + V_2$

$= \frac {1}{ 4 \pi \epsilon_0}[ \frac {q_1}{r} + \frac {q_2}{R} ]$

$= - \frac {1}{ 4 \pi \epsilon_0} [ \frac {Qr}{(R^2 +r^2)} + \frac {QR}{ (R^2 + r^2 ) } ]$

$= \frac {1}{ 4 \pi \epsilon_0} \frac {Q ( R+r) }{ (R^2 +r^2)}$
1581061_1752081_ans_871305a91d85409ba946c543c5658061.jpg

A small uncharged metallic sphere is positioned exactly at a point midway between two equal and opposite point charges separated by very small distance. If the spheres is slightly displaced towards the positive charge and released, then-

0%
It will oscillate about it's original position.
0%
It will move further towards the positive charge.
0%
Its potential energy will decrease and kinetic energy will increase.
0%
The total energy remains constant but is non-zero.

A positively charged particle is released from rest in a uniform electric field. The electric potential energy of the charge

0%
remains a constant because the electric field is uniform.
0%
increases because the charge moves along the electric field.
0%
decreases because the charge moves along the electric field.
0%
decreases because the charge moves opposite to the electric field.

Explanation

Equipotential surface is always perpendicular to the direction of electric field. Positive charge experiences the force in the direction of electric field.

When a positive charge is released from rest in uniform electric field, its velocity increases in the direction of electric field. So K.E. increases, and the P.E. decreases due to law of conservation of energy.

So P.E. of positively charged particle decreases because speed of charged particle moves in the direction of field due to force

$q \bar{E}$

Consider a uniform electric field in the

$\hat{z}$ direction. The potential is a constant

0%
in all space.
0%
for any $x$ for a given z.
0%
for any y for a given z.
0%
on the $x$ -y plane for a given z.

Explanation

As we know that equipotential surfaces are perpendicular to the direction electric field
lines. Here electric field is in $+\hat{z}$ direction.

So, equipotential surfaces will be the plane perpendicular to $z$ axis, i.e.,
along $x$ -y, plane, which includes any $x$ or $y$ axes. So answers $(b),(c)$ and $(d)$ are verified respectively.

In a region of constant potential:

0%
the electric field is uniform
0%
the electric field is zero
0%
there can be no charge inside the region.
0%
the electric field shall necessarily change if a charge is placed outside the region.

Explanation

Constant
potential $\Rightarrow \mathrm{dV}=0$ so by relation $E=\dfrac{-d V}{d r}, \mathrm{E}=0$

i.e., the
E.F. is not uniform discards answer (a) and agree which answer (b).

As potential
may be outside the charge also so there can be no charge inside the region of
constant potential. It verifies answer (c).

If a charge
is placed in outside region, potential difference in region will not be changed
or electric field will not be changed. It makes answer (d) false.

In the figure , the dielectric present in the half part of the plates of a parallel plate capacitor whose dielectric constant is is displaced. If the initial capacitance of the capacitor is C, then the now capacitance will be:

0%
$\dfrac{C}{2}(\varepsilon_r + 1 )$
0%
$\dfrac{1}{2}\dfrac{C}{(\varepsilon _r + 1)}$
0%
$\dfrac{(1 + \varepsilon _r)}{2C}$
0%
$C(1 + \varepsilon _r)$

Explanation

The arrangement shown in the figure is equivalent to the two capacitor joined in parallel. The area of plate in each capacitor is

$A/2$ :

$C_1 = \dfrac{\varepsilon_0A}{2d}$
$$C_2 = \dfrac{\varepsilon _0\epsilon_rA}{2d}$$

$C_{eq} = C_1 + C_2 = \dfrac{\varepsilon _0 A}{2d}[1 + \varepsilon_r]$

$C = \dfrac{\varepsilon _0 A}{d}$

$C_{eq} = \dfrac{C}{2}[1 + \varepsilon_r]$

In a certain region of space, the electric field is zero. From this fact, what can you conclude about the electric potential in this region?

0%
It is zero.
0%
It does not vary with position.
0%
It is positive.
0%
It is negative.

The capacitance of a parallel plate capacitor with air as medium is

$6 \mu F$ . With the introduction of a dielectric medium, the capacitance becomes

$30\mu F$ . The permittivity of the medium is :

$(\in_0= 8.85 \times 10^{-12} C^2N^{-1} m^{-2})$

0%
$1.77 \times 10^{-12} C^2 N^{-1} m^{-2}$
0%
$0.44 \times 10^{-10} C^2 N^{-1} m^{-2}$
0%
$5.00 C^2 N^{-1} m^{-2}$
0%
$0.44 \times 10^{-13} C^2 N^{-1} m^{-2}$

Explanation

Capacitance of air capacitor

$C_0 = \dfrac{\varepsilon_0A}{d} = 6 \mu F$ ......... (i)
When a dielectric of permittivity

$\varepsilon_r$ and dielectric constant K is introduced between the plates, then
Capacitance,

$C = \dfrac{K \varepsilon_0 A}{d} = 30 \mu F$ ..... (ii)
Dividing eq. (ii) by (i), we get

$\dfrac{C}{C_0} = \dfrac{\dfrac{K\epsilon_o A}{ d}}{\dfrac{\varepsilon_0 A}{d}} = \dfrac{30}{6}$

$\Rightarrow K = 5$

$\therefore$ permittivity of the medium

$\varepsilon_0 = \varepsilon_0 K$

$= 8.85 \times 10^{-12} \times 5 = 0.44 \times 10^{-10}$

A capacitor with very large capacitance is in series
with another capacitor with very small capacitance.
What is the equivalent capacitance of the combination?

0%
slightly greater than the capacitance of the
large capacitor
0%
slightly less than the capacitance of
the large capacitor
0%
slightly greater than the capacitance of the small capacitor
0%
slightly less than the capacitance of the small capacitor

A fully charged parallel-plate capacitor remains connected to a battery while you slide a dielectric between the plates. Then the capacitance :

0%
Increase
0%
Decrease
0%
Stay the same
0%
Can not say.

Explanation

Because

$C=k\epsilon \,A/d$ and the dielectric constant

$k$ increases. So C also increases.

A capacitor is filled with an insulator and a certain potential difference is applied to its plates. The energy stored in the capacitor is $U$ . Now the capacitor is disconnected from the source and the insulator is pulled out of the capacitor. The work performed against the forces of the electric field in pulling out the insulator is $4U$ . Then the dielectric constant of the insulator is :

0%
$4$
0%
$8$
0%
$5$
0%
$3$

Explanation

Correct answer: Option C

Hint: A capacitor is defined as an electrical component which stores energy electrostatically in an electric field.

Step 1: Finding final energy

The capacitor originally had the energy U and the work required to pull the insulator out is 4U so the final energy is equal to 5U.

$W + U = {U_t}$

${U_t} = 4U + U$

${U_t} = 5U$

Step 2: Stored energy in terms of capacitance

The capacitance of the insulator with an insulator,

$C = \dfrac{{A{\varepsilon _0}}}{d}k$

The formula of the energy stored in the insulator is given by,

$U = \dfrac{{{q^2}}}{{2C}}$

Where capacitance is C the area of the insulator is A the dielectric constant is k the distance between the plates is d and k is permittivity of dielectric constant.

When we remove the insulator from the capacitor then the capacitance gets ${C_0}$

$U=\dfrac{{{q^2}}}{{2C}}$

${C_0} = \dfrac{{A{\varepsilon _0}}}{d}$

And the stored energy will be,

${U_t} = \dfrac{{{q^2}}}{{2{C_0}}}$

Since,

${U_t} = 5U$

Step 3: Substituting values

Replacing the value of $U$ Ut=5U and ${U_t}$ Uin the above equation we get,

$(\dfrac{{{q^2}}}{{2{C_0}}}) = 5(\dfrac{{{q^2}}}{{2C}})$

$\dfrac{C}{{{C_0}}} = 5$

Replacing the value of $C$ and ${C_0}$ in the above relation

$\dfrac{{\dfrac{{A{\varepsilon _0}}}{d}k}}{{\dfrac{{A{\varepsilon _0}}}{d}}} = 5$

$k = 5$

The dielectric constant value is equal to 5

An air filled parallel plate capacitor has a capacitance

$1 pF$ . The separation between the plates is doubled and wax is inserted between the plates, then the capacitance becomes

$2 pF$ . What is the dielectric constant of wax ?

0%
$8$
0%
$2$
0%
$4$
0%
$16$

Explanation

Capacitance of parallel plate capacitor with air , $C_a=\dfrac{\varepsilon _{0}A}{d} ....(1)$

Capacitance of parallel plate capacitor with wax , $C_w=\dfrac{K\varepsilon _{0}A}{2d} .....(2)$

$(1)/(2),$

$\dfrac{C_a}{C_w}=\dfrac{2}{K}$

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

$\therefore K=4$

‘A’ and ‘B’ are two condensers of capacities 2 $\mu F$ and 4 $\mu F$ They are charged to potential differences of 12V and 6V respectively. If they are now connected (+ve to +ve), the charge that flows through the connecting wire is :

0%
24 $\mu$ C from A to B
0%
8 $\mu$ C from A to B
0%
8 $\mu$ C from B to A
0%
24 $\mu$ C from B to A

Explanation

$Q_A=C_A\ V_A$

$=2\times 10^{-6}\times 12$

$Q_A=24\times 10^{-6}$

$Q_B=C_B\ V_B$

$=4\times 10^{-6}\times 6$

$Q_B=24\times 10^{-6}$

Now when plates are attached total charge

$=Q_A+Q_B$

$Q=48\times 10^{-6}C$
Let x he charge on A then Q-x charge is on B

$\Rightarrow x=2\mu F\times V\ and\ Q-x=4\mu \ F\times V.$

$\Rightarrow \frac{x}{Q-x}=\frac{1}{2}\ \ \ 2x=Q-x\ \ \ \ x=\frac{9}{3}$

$x=16\times 10^{-6};Q-x=32\times 10^{-6}$
Now charge flows from A to B is given by,

$=final -initial$

$=32\times 10^{-6}-20\times 10^{-6}$

$\bigtriangleup Q=8\times 10^{-6} C$

$\bigtriangleup Q = 8\mu C$

Two identical capacitors are connected as shown in the figure. A dielectric slab is introduced between the plates of one of the capacitors so as to fill the gap, the battery remaining connected. The charge on each capacitor will be :(charge on each condenser is $q_{0}$ ; $k$ = dielectric constant )

0%
$\dfrac{2q_{0}}{1+1/k}$
0%
$\dfrac{q_{0}}{1+1/k}$
0%
$\dfrac{2q_{0}}{1+k}$
0%
$\dfrac{q_{0}}{1+k}$

Explanation

When capacitors are connected in series the charge present on them must be equal.

$\dfrac{Q}{c_{1}}+\dfrac{Q}{c_{2}}=2$ ------------ (1)
Now when

$k$ is introduced

$c_{2}$ becomes

$kc_{2}$

Let

$q_{o}$ be the charge on capacitor
given

$c_{1}=c_{2}$

$\Rightarrow 2\dfrac { { q }_{ o } }{ c } =2,\dfrac { { q }_{ o } }{ c } =1$ from (1)

$c ={ q }_{ o }$

After introducing the dielectric, let charge on each plate become

$Q_{ 1 }$

$\Rightarrow \dfrac { Q_{ 1 } }{ { q }_{ o } } +\dfrac { Q_{ 1 } }{ k{ q }_{ 0 } } =2$

$\Rightarrow \dfrac { kQ_{ 1 }+Q_{ 1 } }{ k{ q }_{ o } } =2$

$\Rightarrow Q_{ 1 }=\dfrac { 2kq_{ o } }{ k+1 }$

$\Rightarrow Q_{1}=\dfrac{2q_{o}}{1+ \dfrac{1}{k}}$

A charged particle q is shot towards another charged particle Q which is fixed, with a speed v. It approaches Q upto a closest distance r and then returns. If q was given a speed 2v, the closest distance of approach would be:

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r
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2r
0%
r/2
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r/4

Explanation

As initially q is at infinite distance from Q, thus

$P.E_i=0$

For case 1: the closest distance given is r and velocity of q is v initially.

Applying conservation of energy,

$P.E_i+K.E_i=P.E_f+K.E_f$

$0+ \dfrac{1}{2}mv^2= \dfrac{KQq}{r}+ 0$

$\implies$

$\dfrac{1}{2}mv^2= \dfrac{KQq}{r}$ ................(1)

For case 1: Let the closest distance be r' and velocity of q is 2v initially.

Applying conservation of energy,

$P.E_i+K.E_i=P.E_f+K.E_f$

$0+ \dfrac{1}{2}m(2v)^2= \dfrac{KQq}{r'}+ 0$

$\implies$

$\dfrac{1}{2}m(2v)^2= \dfrac{KQq}{r'}$ ................(2)

Solving (1) and (2),

$r'=\dfrac{r}{4}$

A parallel plate capacitor C with plates of unit area and separation d is filled with a liquid of dielectric constant

$K = 2.$ The level of liquid is

$\dfrac{d}{3}$ initially. Suppose the liquid level decreases at a constant speed V, the time constant as a function of time

$t$ is :

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$\displaystyle \dfrac{6\epsilon_{0}\mathrm{R}}{5\mathrm{d}+3\mathrm{V}\mathrm{t}}$
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$\displaystyle \dfrac{(15\mathrm{d}+9\mathrm{V}\mathrm{t})\epsilon_{0}\mathrm{R}}{2\mathrm{d}^{2}-3\mathrm{d}\mathrm{V}\mathrm{t}-9\mathrm{V}^{2}\mathrm{t}^{2}}$
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$\displaystyle \dfrac{6\epsilon_{0}\mathrm{R}}{5\mathrm{d}-3\mathrm{V}\mathrm{t}}$
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$\displaystyle \dfrac{(15\mathrm{d}-9\mathrm{V}\mathrm{t})\epsilon_{0}\mathrm{R}}{2\mathrm{d}^{2}+3\mathrm{d}\mathrm{V}\mathrm{t}-9\mathrm{V}^{2}\mathrm{t}^{2}}$

Explanation

At time t the level of liquid is,

$\dfrac { d }{ 3 }$ $-Vt\\ { C }_{ eq }=\dfrac { \dfrac { { \epsilon }_{ 0 }KA }{ \dfrac { d }{ 3 } -Vt } \cdot \dfrac { { \epsilon }_{ 0 }A }{ \dfrac { 2d }{ 3 } +Vt } }{ \dfrac { { \epsilon }_{ 0 }KA }{ \dfrac { d }{ 3 } -Vt } +\dfrac { { \epsilon }_{ 0 }A }{ \dfrac { 2d }{ 3 } +Vt } } =\dfrac { { \epsilon }_{ 0 }KA }{ K\left( \dfrac { 2d }{ 3 } +Vt \right) +\dfrac { d }{ 3 } -Vt } \\ =\dfrac { 2{ \epsilon }_{ 0 } }{ \left( \dfrac { 4d }{ 3 } +2Vt \right) +\dfrac { d }{ 3 } -Vt } =\dfrac { 2{ \epsilon }_{ 0 } }{ \left( \dfrac { 5d }{ 3 } +Vt \right) } =\dfrac { 6{ \epsilon }_{ 0 } }{ 5d+3Vt } \\ \tau =R{ C }_{ eq }=\dfrac { 6{ \epsilon }_{ 0 }R }{ 5d+3Vt }$

CBSE Questions for Class 12 Medical Physics Electrostatic Potential And Capacitance Quiz 11 - 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 11 - MCQExams.com

0:0:1

Practice Class 12 Medical Physics Quiz Questions and Answers

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