MCQ Questions for Class 12 Medical Physics Electrostatic Potential And Capacitance Quiz 5

A photographic flash unit consists of a xenon-filled tube. It gives a flash of average power

$2000W$ for

$0.04s$ . The flash due to discharge of a fully charged capacitor of

$40\mu F$ . The voltage to which it is charged before a flash is given by the unit is :

0%
$1500V$
0%
$2000V$
0%
$2500V$
0%
$3000V$

Explanation

Total energy stored in capacitor,

$power\times time=2000\times 0.04J$ (i)

Energy stored in capacitor can also be written as,

$E=\dfrac{CV^2}{2}=\cfrac{1}{2}\times 40\times {10}^{-6} {V}^{2}$ (ii)

using equation (i) and (ii)

$2000\times 0.04=\cfrac{1}{2}\times 40\times {10}^{-6} {V}^{2}$
or

${ V }^{ 2 }=4\times { 10 }^{ -6 }\quad or\quad V=2000V$

An electron is taken from point

$A$ to point

$B$ along the path

$AB$ in a uniform electric field of intensity

$\displaystyle E = 10 Vm^{-1}$ . Side

$AB = 5 m$ , and side

$BC = 3 m$ . Then, the amount of work done on the electron by us is :

0%
50 eV
0%
40 eV
0%
-50 eV
0%
-40 eV

Explanation

$\displaystyle W_{AB} = W_{AC}+W_{CB}$

$\displaystyle W_{CB}$ should be zero, because in moving from

$C$ to

$B$ , we always move perpendicular to field. Hence, force applied by field and displacement will be at

$90^{\circ}$ .
so work done in BC will be 0

$\displaystyle W_{AC} = -e(V_C-V_A)$

$\displaystyle V_C-V_A = -E \times AC = -10 \times 4 = -40$

$\displaystyle \therefore W_{AB} = 40e J=40 eV$

A thin metallic spherical shell contains a charge

$Q$ on its surface. A point charge

$q_{1}$ is placed at the centre of the shell and another charge

$q_{2}$ is placed outside the shell. All the three charges are positive. Then, the force on charge

$q_{1}$ is

0%
Towards left
0%
Towards right
0%
Upward
0%
Zero

Explanation

$\textbf{Step 1: Electrostatic Shielding}$

Due to Electrostatic shielding, The metallic spherical shell acts as a shield and do not allow the electric field due to outside charges to penetrate through it.

Hence, electric field inside shell due to

$Q$ and

$q_1$ is zero.

Therefore, no force will act on charge

$q$ .

Hence, Correct answer is

$(D)$

$\textbf{Note : }$ Electrostatic shielding acts in case of closed metallic surfaces with cavity in which external field does not effect internal contents and internal field does not effect external contents.

Which of the following factors do not influence the capacity of a capacitor?

0%
distance between the plates
0%
material of the plates
0%
area of the plates
0%
curvature of the plates

Explanation

The capacitance of parallel plate capacitor is

$C=\frac{A\epsilon_0}{d}$ where

$A=$ area of plates and

$d=$ distance between plates.
The capacitance also depends on the curvature of plates because spherical plates capacitor and cylindrical plates have different capacitance.
Thus the capacitance is independent of material of plates.

Two identical parallel plate capacitors are connected in series and then joined in series with a battery of

$100V$ . A slab of dielectric constant

$K=3$ is inserted between the plates of the first capacitor. Then, the potential difference across the capacitors will be respectively,

0%
$25V, 75V$
0%
$75V, 25V$
0%
$$20V, 780
0%
$50V, 50V$

Explanation

After inserting a dielectric plate

$C_1=3C and C_2=C$

${ V }_{ 1 }=\cfrac { C\times V }{ 3C+C } =\cfrac{C\times 100}{4C}=25V,\quad { V }_{ 2 }=100-25=75V$

The separation between the plates of

$C_1$ of Fig is doubled and a dielectric of strength 6 is added in between the plates of

$C_2$ . Find the the change in potential drop across

$C_1$ and

$C_2$ :

0%
$\Delta V_1=\dfrac{11C_2V_0C_1}{(C_1+12C_2)(C_1+C_2)}(increases)$

$\Delta V_2=\dfrac{11C_1C_2V_0}{(C_1+C_2)(C_1+12C_2)}(decreases)$
0%
$\Delta V_1=\dfrac{11C_2V_0C_1}{(C_1+12C_2)(C_1+C_2)}(decreases)$

$\Delta V_2=\dfrac{11C_1C_2V_0}{(C_1+C_2)(C_1+12C_2)}(increases)$
0%
$\Delta V_1=\dfrac{22C_2V_0C_1}{(C_1+12C_2)(C_1+C_2)}(increases)$

$\Delta V_2=\dfrac{22C_1C_2V_0}{(C_1+C_2)(C_1+12C_2)}(decreases)$
0%
$\Delta V_1=\dfrac{22C_2V_0C_1}{(C_1+12C_2)(C_1+C_2)}(decreases)$

$\Delta V_2=\dfrac{22C_1C_2V_0}{(C_1+C_2)(C_1+12C_2)}(increases)$

Explanation

$C=\epsilon A/d$

$d_1=2d_0$

$\epsilon_2=6\epsilon_0$

$C'_1=\dfrac{C_1}{2} and C'_2=6C_2$
from circuit of Fig 20.60(a)

$V_1=\dfrac{C_2V_0}{C_1+C_2}\ V_2=\dfrac{C_1V_0}{C_1+C_2}$
From circuit of Fig 20.60 (b)

$V'_1=\dfrac{6C_2V_0}{\dfrac{C_1}{2}+6C_2}=\dfrac{12C_2V_0}{C_1+12C_1};$

$V'_2=\dfrac{\dfrac{C_1}{2}V_0}{\dfrac{C_1}{2}+6C_2}=\dfrac{C_1V_0}{C_1+12C_2}$

$\Delta V_1=\dfrac{12C_2V_0}{C_1+12C_2}-\dfrac{C_2V_0}{C_1+C_2}$

$=C_2V_0\left [ \dfrac{12(C_1+C_2)-(C_1+12C_2)}{(C_1+12C_2)(C_1+C_2)} \right ]$

$=\dfrac{11C_2V_0C_1}{(C_1+12C_2)(C_1+C_2)}(increases)$

$\Delta V_2=\dfrac{C_1V_0}{C_1+C_2}-\dfrac{C_1V_0}{C_1+12C_2}$

$=C_1V_0\left ( \dfrac{C_1+12C_2-(C_1+C_2)}{(C_1+C_2)(C_1+12C_2)} \right )$

$=\dfrac{11C_1C_2V_0}{(C_1+C_2)(C_1+12C_2)}(decreases)$

A hollow insulated conducting sphere is given a positive charge of

$10\mu C$ . What will be the electric field at the centre of the sphere if its radius is

$2$ metres?

0%
zero
0%
$5\mu C{m}^{-2}$
0%
$20\mu C{m}^{-2}$
0%
$8\mu C{m}^{-2}$

Explanation

$\textbf{Step 1: Uniform Charge distribution on outer surface [Refer Figure]}$

As a property of conductor, Charge will resides on the outer surface of the hollow sphere.

Since the shape of sphere is symmetric, hence charge distribution will be uniform, As shown in the figure.

$\textbf{Step 2: Finding Electric field inside}$

The given situation is now a uniformly charged hollow conducting shell.

Inside which the Electric field is zero at all points.

Hence correct option is

$A$ .

$\large \textbf{Alternate Solution using Gauss Law:}$

We can find the electric field inside using Gauss Law as follows:

Consider a gaussian spherical surface of radius

$r<R$

Charge inside gaussian surface

$q_{in}=0$

From gauss theorem:

$\oint _{ }^{ }{ \vec { E } .\vec {ds} }$

$=\cfrac { q_{in} }{ { \varepsilon }_{ 0 } }$

At all points of gaussian surface,

$\vec E$ is constant radially outwards due to symmetry and

$\vec {ds}$ is perpendicular to the gaussian surface, hence both are parallel and angle between them is zero.

$\therefore\ \ \ \vec E \oint _{ }^{ }{ \vec {ds}\ cos0^{o}}$

$= \vec E \oint \vec {ds}$

$=\cfrac { q_{in} }{ { \varepsilon }_{ 0 } }$

Full area of gaussian surface is

$4 \pi r^2$

$\Rightarrow\ \ \ \ E\times 4\pi { r }^{ 2 }=\cfrac { 1 }{ { \varepsilon }_{ 0 } } \times 0\quad$

$\Rightarrow \quad E=0$

Hence, option A is correct.

2107848_195748_ans_4f2dc7b1f4c946bcba8422ea5199b720.png

A parallel plate capacitor is charged to a certain voltage. Now, if the dielectric material (with dielectric constant k) is removed then the

0%
capacitance increases by a factor of k
0%
electric field reduces by a factor k
0%
voltage across the capacitor decreases by a factor k
0%
none of these

Explanation

As the capacitor is charged by using cell so potential as well as filed between the plates become constant.
For removing dielectric the capacitance becomes

$C/k$ . Thus capacitance decreases by a factor of k.

A foil of aluminium of negligible thickness is inserted in between the space of a parallel plate condenser. If the foil is electrically insulated, the capacity of the condenser will :

0%
increase
0%
decrease
0%
remain unchanged
0%
become zero

Explanation

The capacity of condenser before inserting foil is

$C=\dfrac{A\epsilon_0}{d}$ where A be the area of plate and

$d$ be the separation between plates.
After inserting foil the there will be two capacitors in series with capacitance 2C as distance is halved and the series combination of the two will give equivalent capacitance of C, hence, capacity will remain same.

For a given potential difference V how would you connect two capacitors, to obtain greater stored charge :

0%
series
0%
parallel
0%
series and parallel combined
0%
cannot say

Explanation

Since

$q=CV$ & capacitance of two capacitor is more in parallel combination as compared to series combinations i.e.,

$C_P > C_S$ .

If in a parallel plate capacitor, which is connected to a battery, we fill dielectrics in whole space of its plates, then which of the following increases?

0%
Q and V
0%
V and E
0%
E and C
0%
Q and C

In a charged capacitor, the energy is stored in :

0%
the negative charges
0%
the positive charges
0%
the field between the plates
0%
both 'a' and 'b'

A positive point charge q is carried from a point B to a point A in the electric field of a point charge

$+$ Q held at O. If the permittivity of free space is

$\epsilon_0$ and OA=a and OB=b, the work done in the process is given by :

0%

$\dfrac {qQ}{4\pi \epsilon_0}\left (\dfrac {1}{a}+\dfrac {1}{b}\right )$
0%

$\dfrac {qQ}{4\pi \epsilon_0}\left (\dfrac {1}{a}-\dfrac {1}{b}\right )$
0%

$\dfrac {qQ}{4\pi \epsilon_0}\left (\dfrac {1}{a^2}-\dfrac {1}{b^2}\right )$
0%

$\dfrac {qQ}{4\pi \epsilon_0}\left (\dfrac {1}{a^2}+\dfrac {1}{b^2}\right )$

Explanation

We know that work done $dW=qdV$

here , $\displaystyle W_{B\rightarrow A}=q{V_A-V_B}=q\left[\dfrac{Q}{4\pi\epsilon_0 (OA)}-\dfrac{Q}{4\pi\epsilon_0(OB)}\right]=\dfrac{qQ}{4\pi\epsilon_0}[\dfrac{1}{a}-\dfrac{1}{b}]$ where $OA=a, OB=b$

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 :

0%
remain same
0%
become zero
0%
increase
0%
decrease

Explanation

The electric field in between two parallel metal plate is

$E=\dfrac{\sigma}{\epsilon}=\dfrac{Q}{A\epsilon}$ where

$A=$ area of plates and

$\epsilon=$ permeability of medium in between the plates.
The permeability of kerosene is more that of air. As

$\epsilon$ increases,

$E$ decreases.

If the potential of a capacitor having capacity of

$6\mu F$ is increased from 10 V to 20 V, then increase in its energy will be :

0%
$4\times 10^{-4}J$
0%
$5\times 10^{-4}J$
0%
$9\times 10^{-4}J$
0%
$12\times 10^{-6}J$

Explanation

Initial energy

$U_i=\frac{1}{2}CV^2_i$ and final energy

$U_f=\dfrac{1}{2}CV^2_f$
Increase in energy

$=\Delta U=U_f-U_i=\dfrac{1}{2}C(V_f^2-V_i^2)$

$=\dfrac{1}{2}\times (6\times 10^{-6})\times [20^2-10^2]$

$=9\times 10^{-4} J$

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

0%
decreases
0%
remains unchanged
0%
becomes infinite
0%
increases

Explanation

Capacitance of parellel plate capacitor with a dielectric slab of thickness t inside is given by: $C=\dfrac{A\epsilon_0}{(d-t)+\dfrac{t}{K}}$ , where K is the dielectric constant.
When aluminium sheet of negligible thickness is inserted, then thickness of slab $t\rightarrow 0$ and value of $K$ for metal (conductor) will be $\infty$
Therefore $C=\dfrac{A\epsilon_0}{d}$
Hence, capacitance of capacitor will remains unchanged.
Option B is correct.

2112080_201396_ans_a88691882ace48f7aa7bf7ad179b4ce2.jpg

A parallel plate condenser is immersed in an oil of dielectric constantThe field between the plates is :

0%
increased, proportional to $\frac {1}{2}$
0%
decreased, proportional to -2
0%
increased, proportional to -2
0%
decreased, proportional to $-\frac {1}{2}$

Two capacitors of capacitance

$C_1$ and

$C_2$ are connected in parallel across a battery. If

$Q_1$ and

$Q_2$ respectively be the charges on the capacitors, then

$\dfrac {Q_1}{Q_2}$ will be equal to :

0%
$\dfrac {C_2}{C_1}$
0%
$\dfrac {C_1}{C_2}$
0%
$\dfrac {C_1^2}{C_2^2}$
0%
$\dfrac {C_2^2}{C_1^2}$

Explanation

In parallel combination the both capacitors have same potential , V (say).
So,

$Q_1=C_1V$ and

$Q_2=C_2V$

$\therefore \dfrac{Q_1}{Q_2}=\dfrac{C_1V}{C_2V}=\dfrac{C_1}{C_2}$

The work done in placing a charge of

$8\times 10^{-18}$ coulomb on a condenser of capacity 100 micro-farad is :

0%
$3.1\times 10^{-26}J$
0%
$4\times 10^{-10} J$
0%
$32\times 10^{-32} J$
0%
$16\times 10^{-232} J$

Explanation

Given , charge

$=q=8\times 10^{-18} C$ and Capacitance

$=C=100 \mu F$
Here the work done is the energy stored in the capacitor. i.e,

$\displaystyle W=\frac{q^2}{C}=\frac{(8\times 10^{-18})^2}{100\times 10^{-6}}=32\times 10^{-32} J$

A parallel plate condenser with oil between the plates (dielectric constant of oil

$K=2)$ has a capacitance C. If the oil is removed, then capacitance of the capacitor becomes :

0%
$\sqrt 2C$
0%
$2C$
0%
$\dfrac {C}{\sqrt 2}$
0%
$\dfrac {C}{2}$

Explanation

When oil is present, the capacitance

$\displaystyle C=\frac{AK\epsilon_0}{d}=\frac{2A\epsilon_0}{d}$ where

$A=$ area of plates and

$d=$ separation of plates and dielectric constant

$K=2$
When oil is removed, the capacitance becomes

$\displaystyle C'=\frac{A\epsilon_0}{d}=\frac{C}{2}$

Two metal spheres of radii a and b are connected by a thin wire. Their separation is very large compared to their dimensions. The capacitance of this system is :

0%
$4\pi \varepsilon_o(ab)$
0%
$2\pi \varepsilon_o(a+b)$
0%
$4\pi \varepsilon_o(a+b)$
0%
$4\pi \varepsilon_o\left (\dfrac {a^2+b^2}{2}\right )$

Explanation

The capacitance of single sphere capacitor is

$C=4\pi\epsilon_0 r$
Thus, the capacitance of sphere of radii a is

$C_a=4\pi\epsilon_0 a$
and the capacitance of sphere of radii b is

$C_b=4\pi\epsilon_0 b$
When they is connected by a wire, they are connected in parallel.
So the capacitance of the system,

$C_{eq}=C_a+C_b=4\pi\epsilon_0 a+4\pi\epsilon_0 b=4\pi\epsilon_0 (a+b)$

Two capacitors of capacitance C are connected in series. If one of them is filled with dielectric substance k, what is the effective capacitance?

0%
$\frac {kC}{(1+k)}$
0%
C(k+1)
0%
$\frac {2kC}{1+k}$
0%
none of these

Explanation

When capacitance filled with dielectric , the capacitance becomes

$C'=kC$
As they are in series so effective capacitance is

$\displaystyle C_{ef}=\frac{CC'}{C+C'}=\frac{C(kC)}{C+kC}=\frac{kC}{1+k}$

Find the dimensions of capacitance :

0%
$=[M^{-1}L^{2}T^4A^{-2}]$
0%
$=[M^{-1}L^{-2}T^{-4}A^2]$
0%
$=[M^{-1}L^{-1}T^4A^2]$
0%
$=[M^{-1}L^{-2}T^4A^2]$

Explanation

From the expression of energy stored in capacitor,

$U=\dfrac {1}{2}\dfrac {q^2}{C}$

$\therefore [C] = \left [\dfrac {q^2}{U} = \right ]\left [\dfrac {A^2T^2}{ML^2T{^-2}}\right ]$

$=[M^{-1}L^{-2}T^4A^2]$

No charge will flow when two conductors having the same charge are connected to each other.

Given statement is:

0%
True
0%
False

The electric potential inside a conductor:

0%
is zero
0%
increases with distance from center
0%
is constant
0%
decreases with distance from center

Explanation

As the electric field inside a conductor is zero so the potential at any point is constant.

Suppose

$a$ and

$b$ two points inside a conductor. we know that

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

$E=0$ ,

$dV=0$ or

$V_a-V_b=0$ or

$V_a=V_b$

If a capacitor

$900\mu F$ is charged to

$100 V$ and its total energy is transferred to a capacitor of capacitance

$100 \mu F$ then its potential is :

0%
$200 V$
0%
$30 V$
0%
$300 V$
0%
$400 V$

Explanation

Let the final potential is

$V_2$ .
As the total energy of first capacitor is transferred to second capacitor so

$U_1=U_2$
Thus,

$\dfrac{1}{2}C_1V_1^2=\dfrac{1}{2}C_2V_2^2$

$900\times 10^{-6}\times (100)^2=100\times 10^{-6}\times V_2^2$

$V_2^2=900\times 100 \Rightarrow V_2=300 V$

If a slab of insulating material

$4\times 10^{-5}m$ thick is introduced between the plates of a parallel plate capacitor, the distance between the plates has to be increased by

$3.5\times 10^{-5}m$ to restore the capacity to original value. Then the dielectric constant of the material of slab is :

0%
$8$
0%
$6$
0%
$12$
0%
$10$

Explanation

$K$ be the dielectric constant, the distance increased due to introduce dielectric is

$x=t-\dfrac{t}{K}=t\left (1-\dfrac{1}{K}\right )$

where

$t=$ thickness of dielectric.
According to question,

$x=3.5 \times 10^{-5}$

$\Rightarrow t\left (1-\dfrac{1}{K}\right )=3.5 \times 10^{-5}$

$\Rightarrow 1-\dfrac{1}{K}=\dfrac{3.5 \times 10^{-5}}{4\times 10^{-5}}=0.875$

$\Rightarrow \dfrac{1}{K}=1-0.875=0.125$

$\Rightarrow K=8$

If we increase '

$d$ ' of a parallel condenser to '

$2d$ ' and fill wax to the whole empty space between its two plate, then capacitance increase from

$1pF$ to

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

0%
$2$
0%
$4$
0%
$6$
0%
$8$

Explanation

Initially the capacitance,

$C_1=\dfrac{A\epsilon_0}{d}$ where

$A=$ area of plates
Final capacitance,

$C_2=\dfrac{AK\epsilon_0}{2d}$ where

$K=$ dielectric constant.

$\therefore \dfrac{C_1}{C_2}=\dfrac{2}{K} \Rightarrow \dfrac{1}{2}=\dfrac{2}{K}$

$\therefore K=4$

An air capacitor C connected to a battery of e.m.f. V acquires a charge q and energy E. The capacitor is disconnected from the battery and a dielectric slab is placed between the plates. Which of the following statements is correct?

0%
V and q decrease but C and E increase
0%
V remains unchange, but q, E and C increase
0%
q remains unchanged, C increases, V and E decrease
0%
q and C increase but V and E decrease

Explanation

As the capacitor is disconnected from battery so charge q will remain uncharged.
When dielectric slab placed, the capacitance becomes

$C'=kC$ . where

$k$ is dielectric constant. Thus the capacitance C increases k times.
As

$V=\frac{q}{C}$ , V decreases because q is constant and C increases.
As

$V=Ed$ , so E also decreases because V decreases.

A bullet of mass

$2 g$ . is having a charge of

$2\:\mu C$ . Through what potential difference must it be accelerated, starting from rest, to acquire a speed of

$10\:m/s$ ?

0%
$5 MV$
0%
$5V$
0%
$50 V$
0%
$5 kV$

Explanation

Here change in kinetic energy

$=$ work done

$=$ potential energy

$\dfrac{1}{2}mv^2=qV_{ac}$

$V_{ac}=\dfrac{mv^2}{2q}=\dfrac{(2\times 10^{-3})\times 10^2}{2\times (2\times 10^{-6})}=5 MV$

In moving from

$A$ to

$B$ along an electric field line, the electric field does

$1.28\times {10}^{-18}J$ of work on an electron. If

${\phi}_{1}, {\phi}_{2}$ are equipotential surface, then potential difference

${V}_{E}-{V}_{D}$ is equal to :

0%
$-8$ volt
0%
$+8$ volt
0%
zero
0%
$64$ volt

Explanation

Work done by field $W=-\Delta U={U}_{A}-{U}_{B}=(-e){V}_{A}-(-e){V}_{B}$

$\Rightarrow$

$1.28\times {10}^{-18}=1.6\times {!0}^{-19}({V}_{B}-{V}_{A})$

${V}_{B}-{V}_{A}=8$ volt

$\Rightarrow$

${V}_{E}-{V}_{D}={V}_{A}-{V}_{B}=-8$ volt

A capacitor becomes a perfect insulator for

0%
alternating current
0%
direct current
0%
both (a) and (b)
0%
None of these

The associated figure shows an electric charge

$+q$ located at the centre of a hollow uncharged conducting metal sphere. Outside the sphere is a second charge

$+Q$ . Both charges are positive. Choose the description below that describes the net electrical forces on each charge in this situation.

0%
$Both\ charges \ experience \ the \ same \ net \ force \ directed \ away \ from \ each \ other.$
0%
$No \ net \ force \ experienced \ by \ either \ charge.$
0%
$There \ is \ no \ net \ force \ on \ Q \ but \ a \ net \ force \ on \ q.$
0%
$There \ is \ no \ net \ force \ on \ q \ but \ a \ net \ force \ on \ Q.$
0%
$Both \ charges \ experience \ a \ net \ force \ but \ they \ are \ different \ from \ each \ other.$

Explanation

We know that a conducting metallic shell acts as shield from outside electric field and there can't be electric field inside a shell due to outside charge.

Hence, there will be no force experienced by charge

$q$ .

But

$Q$ will experience force according to Coulomb's law.

Answer-(D)

A point charge

$q_1 = q$ is placed at point P. Another point charge

$q_2 = - q$ is placed at point Q. At some point

$R(R \neq p, R \neq Q)$ electric potential due to

$q_1$ is

$V_1$ and electric potential due to

$q_2$ is

$V_2$ . Which of the following is correct?

0%
Only for some points $V_1 > V_2$ ,
0%
Only for some points $V_2 > V_1$ ,
0%
For all points $V_1 > V_2$
0%
For all points $V_2 > V_1$

Explanation

Potential at R due to point charge at P,

$V_1=\dfrac{kq_1}{x}=\dfrac{kq}{x}$
where x is distance between point P and R.
Potential at R due to point charge at Q,

$V_2=\dfrac{kq_2}{y}=\dfrac{k(-q)}{y}$
where x is distance between point Q and R.
If the charge is positive, potential is positive and if the charge is negative, potential is negative. Hence option c is correct.

An electrical charge

$2 \times 10^{-8}$ C is placed at the point (1, 2, 4 )m. At the point (4, 2, 0) m, the electric

0%
potential will be 36 V
0%
field will be along y-axis
0%
field will increase if the space between the points is filled with a dielectric
0%
All of the above

Explanation

Here, charge

$q=2\times 10^{-8} C$ and

$r=\sqrt{(1-4)^2+(2-2)^2+(4-0)^2}=\sqrt{25}=5 m$

and

$\hat{r}=\dfrac{\vec{r}}{r}=\dfrac{1}{5}[(1-4)\hat i+(2-2)\hat j+(4-0)\hat k]=\dfrac{1}{5}[-3\hat i+4\hat k]$

Thus, potential

$V=\dfrac{1}{4\pi\epsilon_0} \dfrac{q}{r}=9\times 10^9 \times \dfrac{2\times 10^{-8}}{5}=36 V$

Electric filed

$\vec{E}=\dfrac{1}{4\pi\epsilon_0} \dfrac{q}{r^2} \hat r=9\times 10^9 \times \dfrac{2\times 10^{-8}}{5^2}. \dfrac{1}{5}[-3\hat i+4\hat k]=1.44[-3\hat i+4\hat k]$

Thus, the field will along

$x-z$ plane.

If space between points is filled by dielectric, the filed becomes

$E'=E/k$ where

$k=$ dielectric constant. since,

$k>1$ so filed will decrease.

A charge of 5 C is moved between two points in an electric field and 20 J of work was done to do so. Calculate the potential difference between the two points

0%
4 V
0%
2 V
0%
3 V
0%
6 V

Explanation

Given:

charge to be moved in an electric field

$=5C=q$

work done

$=20J$

we have to find potential difference b/w 2 point.

here, we use the formula that

$work done =(charge\times potential\quad difference)$ b/w 2 point.

so, potential difference b//w 2 point

$V=\cfrac { work\quad done }{ charge } \\ =\cfrac { 20 }{ 5 } =4V$ .

Let S be the set of all points in a plane. Let R be a relation on S such that for any two points a and b, aRb iff b is within 1 cm from a. Then R is

0%
Equivalence relation
0%
Reflexive and symmetric not transitive
0%
Reflexive and transitive not symmetric
0%
Symmetric and transitive not reflexive

Explanation

R is reflexive since any point is at distance O from itself so that it is within 1 cm from itselt.
R is symmetric since if a point a is within 1 cm from another point b, then b is also within 1 cm from a.
R is not transitive. Let a, b, c, be three points in a straight lines in this order such that distance between a and b is

$\frac{1}{2}$ cm and distance between band c is

$\frac{2}{3}$ cm. Then aRb and bRc. But (a, c) distance between a

$\not{\varepsilon}$ R since the and c is

$\displaystyle \left ( \frac{1}{2} + \frac{2}{3} \right ) = \frac{7}{6} cm > 1 cm.$

Which of the following is not a unit for electric potential?

0%
Volt
0%
Joule/coulomb
0%
Erg/stat coulomb
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None of these

Explanation

Volt is SI unit of electric potential.
Also,

$V=\dfrac{Work}{Charge}=\dfrac{J}{Coulomb}$

In C.G.S, Volt

$=\dfrac{Erg}{Stat\;Coulomb}$

So, a, b, c are units of potential.

$ECG$ is a method of diagnosis used to check the functioning of heart, it is based on the principle that there is

0%
A pattern of magnetic field associated with the functioning of heart
0%
An electrical impulse associated with the functioning of organs and tissues of human body
0%
A varying electrical field associated. with organs and tissues of human body
0%
A varying magnetic field associated with organs and tissues of human body

The electric potential at the surface of an atomic nucleus (Z = 50) of radius

$9.0 \times 10^{-13} cm$ is:

0%
$9\times 10^5$ volt
0%
$8 \times 10^6$ volt
0%
80 volt
0%
9 volt

Explanation

Electric potential at the surface of an atomic nucleus,

$V = k. \displaystyle \dfrac{Ze}{r}$

$=\displaystyle \frac{9\times 10^9 \times 50 \times 1.6 \times 10^{-19}}{9 \times 10^{-13} \times 10^{-2}} = 8 \times 10^6 V$

what is the potential difference between two points, if 2J of work must be done to move a 4 mC charge from one point to another is:

0%
50 V
0%
500 V
0%
5 V
0%
5000 V

Identify the correct statement from below:

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A will be at higher potential
0%
B will be at higher potential
0%
Both will be at same potential
0%
Can't be said

Assertion: A parallel plate capacitor is connected across battery through a key. A dielectric slab of dielectric constant K is introduced between the plates. The energy which is stored becomes K times.
Reason: The surface density of charge on the plate remains constant or unchanged.

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If both assertion and reason are true but the reason is the correct explanation of assertion.
0%
If both assertion and reason are true but the reason is not the correct explanation of assertion.
0%
If assertion is true but reason is false.
0%
If both the assertion and reason are false.
0%
If reason is true but assertion is false.

Explanation

Energy stored in the capacitor

$U = \dfrac{1}{2}CV^2$

When a dielectric slab of dielectric constant K is introduced between the plates of the condenser, then

it becomes KC

So energy stored will become K times.
Since Q =CV , So Q will become K times

∴ Surface charge density

$\sigma = \dfrac{KQ}{A} K\sigma_o$

Assertion: If the distance between parallel plates of a capacitor is halved and dielectric constant is three times, then the capacitance becomes 6 times.
Reason : Capacity of the capacitor does not depend upon the nature of the material.

0%
If both assertion and reason are true but the reason is the correct explanation of assertion.
0%
If both assertion and reason are true but the reason is not the correct explanation of assertion.
0%
If assertion is true but reason is false.
0%
If both the assertion and reason are false.
0%
If reason is true but assertion is false.

Among identical spheres A and B having charges

$-15\ C$ and

$-16\ C$ :

0%
$-15\ C$ is at higher potential
0%
$-16\ C$ is at higher potential
0%
both are at equal potential
0%
no such comparison can be made

Explanation

Potential at surface of a sphere of radius r is

$V=kQ/r$

As identical sphere so r is same for both, Thus V will depend on charge Q.

So when Q is more , potential will be more.

Since

$-15\ C$ is more than

$-16\ C$ , so

$-15\ C$ sphere will have higher potential.

A parallel plate air capacitor has a capacitance

$C$ .When it is half filled with a dielectric of dielectric constant

$5$ , the percentage increase in the capacitance will be :

0%
$400$ %
0%
$66.6$ %
0%
$33.3$ $ %
0%
$200$ %

Explanation

Initial capacitance

$= \dfrac{\epsilon_oA}{d}$

When it is half filIed by a dielectric of dielectric constant K, then

$C_1 = \dfrac{K\epsilon_oA}{d/2} = 2K \dfrac{\epsilon_oA}{d}$

and

$C_2 = \dfrac{\epsilon_oA}{d/2} = \dfrac{2\epsilon_oA}{d}$

$\dfrac{1}{C} = \dfrac{1}{C_1} + \dfrac{1}{C_2} = \dfrac{d}{2\epsilon_oA}(\dfrac{1}{K} + 1) = \dfrac{d}{2\epsilon_oA}(\dfrac{1}{5}+1) = \dfrac{3}{5} \dfrac{d}{\epsilon_oA}$

$C = \dfrac{5}{3} \dfrac{\epsilon_oA}{d}$

Hence % increase in capacitance

$\bigg (\dfrac{\dfrac{5\epsilon_oA}{d}-\dfrac{\epsilon_oA}{d}}{\dfrac{\epsilon_oA}{d}}\bigg ) \times 100 = \dfrac{2}{3} \times 100 = 66.6$ %

A capacitor is charged by using a battery which is then disconnected. A dielectric slab is introduced between the plates which results in :

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increase in the potential difference across the plates and reduction in stored energy but no change in the charge on the plates
0%
decrease in the potential difference across the plates and reduction in the stored energy but no change in the charge on the plates
0%
reduction of charge on the plates and Increase of potential difference across the plates
0%
increase in stored energy but no change in potential difference across the plates

Explanation

If a dielectric slab of dielectric constant K is filled in between the plates of a capacitor after charging the capacitor (i.e., after removing the connection of battery with the plates of capacitor) the potential difference between the plates reduces to

$\dfrac{1}{K}$ times and the potential energy of capacitor reduces to

$\dfrac{1}{K}$ .

times but there is no change in the charge on the plates.

Three condenser of capacitance

$C(\mu F)$ are connected in parallel to which a condenser of capacitance

$C$ is connected in series. Effective capacitance is

$3.75$ , then capacity of each condenser is

0%
$4\mu F$
0%
$5\mu F$
0%
$6\mu F$
0%
$8\mu F$

Explanation

The effective capacitance of three condenser connected in parallel

$=3C$ .
When

$3C$ is connected in series to

$C$

$C_{Result}=\displaystyle\frac{3C\times C}{3C+C}=3.75$

$\Rightarrow C=5\mu F$ .

"The work per unit of charge required to move a charge from a reference point to a specified point, measured in joules per coulomb or volts. The static electric field is the negative of the gradient of the electric potential." comments are given below ,select the correct one

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Statement is correct
0%
Statement is incorrect
0%
Comment on electric field is not correct
0%
Only comment on electric potential is not correct

Explanation

The given statement is correct. Volt or Joules per Coulomb are SI unit of Potential (V) i.e, work done per unit charge to move a charge from a reference point to a certain point.
And electric field E is negative of the gradient of electric potential (V) i.e,

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

Three particles, each having a charge of

$10\;\mu C$ are placed at the corners of an equilateral triangle of side

$10\ cm$ . The electronic potential energy of the system is: (Given

$\dfrac{1}{4\pi \varepsilon_0}=9\times10^9N-m^2/C^2$ )

0%
$Zero$
0%
Infinite
0%
$27\;J$
0%
$100\;J$

Explanation

Three particles are situated at A,B and C having charge

$Q = 10 \mu C$ each.

The distance between the two charges

$r = 10 cm = 0.1$

$m$

Potential energy of the charges at A and B,

$E_{AB} = \dfrac{1}{4\pi \epsilon_o} \dfrac{Q^2}{r}$

$\therefore$

$E_{AB} = \dfrac{9 \times 10^9 \times (10 \times10^{-6})^2 }{0.1} = 9 J$

Similarly,

$E_{BC} = E_{AC} = 9J$

Total potential energy of the system,

$E = E_{AB} + E_{BC} +E_{AC} =27 J$

CBSE Questions for Class 12 Medical Physics Electrostatic Potential And Capacitance Quiz 5 - 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 5 - MCQExams.com

0:0:1

Practice Class 12 Medical Physics Quiz Questions and Answers

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