Electrostatic Potential and Capacitance Test

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Electrostatic Potential and Capacitance Test - 47

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Weekly Quiz Competition

Question 1
1 / -0

Three capacitors of capacitance $1.0,2.0$ and $5.0\mu F$ are connected in series to a $10V$ source. The potential difference across the $2.0\mu F$ capacitor is

A
$\cfrac { 100 }{ 17 }$

B
$\cfrac { 20 }{ 17 } V$

C
$\cfrac { 50 }{ 17 } V$

D
$10V$

Solution

${ C }_{ eq }=\cfrac { 10 }{ 17 } \mu F,Q=\cfrac { 100 }{ 17 } \mu C\quad$
$\therefore$ Potential difference across $2\mu F$ capacitor
$=\cfrac { \cfrac { 100 }{ 17 } \mu C }{ 2\mu F } =\cfrac { 50 }{ 17 } V$
Question 2
1 / -0

A metal plate of thickness $d/2$ is introduced in between the plates of a parallel plate air capacitor with plate separation of $d$ . Capacity of the metal plate :

A
Decreases $2$ times

B
Increases $2$ times

C
Remains same

D
Becomes zero

Solution

$C=\dfrac{\epsilon_oA}{d}$
when a dielectric is filled ,thickness t
$C_1=\dfrac{\epsilon_oA}{d-t+\frac{t}{K}}$
for metal k= infinity , $t=\dfrac{d}{2}$
$C_1=\dfrac{2\epsilon_oA}{d}=2C$
Question 3
1 / -0

An electron enters into a space between the plates of parallel plate capacitor at an angle of $\alpha$ with the plates and leaves at an angle of $\beta$ to the plates. The ratio of its $KE$ while leaving to entering the capacitor will be :

A
$\left (\dfrac {\cos \alpha}{\cos \beta}\right )^{2}$

B
$\left (\dfrac {\cos \beta}{\cos \alpha}\right )^{2}$

C
$\left (\dfrac {\sin \alpha}{\sin \beta}\right )^{2}$

D
$\left (\dfrac {\sin \beta}{\sin \alpha}\right )^{2}$

Solution

We know that Electric field is always perpendicular to the plates of the capacitor.

Let the speed of the electron be $v_1$ when it enters the gap between the plates and $v_2$ when it leaves the plates.

As the electron only experiences force in the direction perpendicular to the plates, by the conservation of momentum in the direction parallel to the plates, we have
$v_1\textrm{cos}(\alpha) = v_2\textrm{cos}(\beta) \Rightarrow \dfrac{v_2}{v_1} = \dfrac{\textrm{cos}\alpha}{\textrm{cos}\beta}$

The ratio of kinetic energy is given by $(\dfrac{v_2}{v_1})^2 = (\dfrac{\textrm{cos}\alpha}{\textrm{cos}\beta})^2$
Question 4
1 / -0

Given below are three schematic graphs of potential energy $V(r)$ versus distance $r$ for three atomic particles: electron $(e^{-})$ , proton $(p^{+})$ and neutron $(n)$ , in the presence of a nucleus at the origin $O$ . The radius of the nucleus is $r_{0}$ . The scale on the V-axis may not be the same for all figures. The correct pairing of each graph with the corresponding atomic particle is.

A
$(1, n), (2, p^{+}), (3, e^{-})$

B
$(1, p^{+}), (2, e^{-}), (3, n)$

C
$(1, e^{-}), (2, p^{+}), (3, n)$

D
$(1, p^{+}), (2, n), (3, e^{-})$

Solution

The interaction forces between the sub-atomic particles are
(i) Electrostatic forces and (ii) Nuclear forces/Strong forces.

The nuclear forces are very large in magnitude compared to electrostatic forces, but they are short distance forces. They only act within the nucleus.

Thus, within the nucleus, the potential is governed by nuclear forces and outside the nucleus, it is governed by the electrostatic forces.

$\textrm{Case 1: Neutron}$
Neutron is a chargeless particle. Thus, outside the nucleus, electrostatic potential energy of neutron is zero.
But, within the nucleus, the nuclear forces (attractive) bind the neutron to the nucleus. Hence potential energy is negative.
Thus, Potential energy of neutron corresponds to figure (1)

$\textrm{Case 2: Proton}$
proton is a positively charged particle. Thus, outside the nucleus, electrostatic forces on the proton are repulsive. Thus, potential energy of proton is positive.
But, within the nucleus, the nuclear forces (attractive) bind the proton to the nucleus. Hence potential energy is negative.
Thus, Potential energy of proton corresponds to figure (2)

$\textrm{Case 3: Electron}$
Electron is a negatively charged particle. Thus, outside the nucleus, electrostatic forces on the electron are attractive. Thus, potential energy of electron is negative.
But, within the nucleus, the nuclear forces and the electrostatic forces are of the same order (since electron is not a intra nuclear particle).
Thus, Potential energy of electron corresponds to figure (3)
Question 5
1 / -0

How many $6\mu F$ , $200 V$ condensers are needed to make a condenser of $18\mu F$ , $600 V$ ?

A
$9$

B
$18$

C
$3$

D
$27$

Solution

Place three 200V, $6\mu F$ capacitors in series to get 1 equivalent 600V, $2\mu F$ capacitor. Now place 9 of these equivalent 600V, $2\mu F$ capacitors in parallel to obtain an equivalence of $18\mu F$ at 600 Volts. All this requires a total of 27 $6\mu F$ capacitors. Nine rows connected in parallel with 3 capacitors connected in series in each row.
Question 6
1 / -0

The minimum value of effective capacitance that can be obtained by combining $3$ capacitors of capacitances $1 pF$ , $2 pF$ and $4 pF$ is

A
$\dfrac { 7 }{ 4 } pF$

B
$\dfrac { 4 }{ 7 } pF$

C
$1 pF$

D
$2 pF$

Solution

Minimum capacitance is obtained when these three capacitors are connected in series with each other.
Effective capacitance of series combination $\dfrac{1}{C_{eff}} = \dfrac{1}{C_1} + \dfrac{1}{C_2} + \dfrac{1}{C_3}$
$\therefore$ $\dfrac{1}{C_{eff}} = \dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{4}$
$\implies$ $C_{eff} = \dfrac{4}{7}p F$
Question 7
1 / -0

Two identical parallel plate capacitance C each are connected in series with a battery of emf, E as shown, If one of the capacitors is now filled with a dielectric of dielectric constant k, the amount of charge which will flow through the battery is $?$ (neglect internal resistance of the battery)

A
$\displaystyle\frac{k+1}{2(k-1)}CE$

B
$\displaystyle\frac{k-1}{2(k+1)}CE$

C
$\displaystyle\frac{k-2}{k+2}CE$

D
$\displaystyle\frac{k+2}{k-2}CE$

Solution

Inital charge on both the capacitor $Q=\dfrac{CE}{2}$
final charge $Q_f=\dfrac{kCE}{K+1}$
charge flow from batter is $Q_f-Q=\dfrac{k-1}{2(k+1)}CE$
Question 8
1 / -0

A charge $10nC$ is situated in a medium of relative permittivity $10$ . The potential due to this charge at a distance of $0.1 m$ is :

A
$900V$

B
$90V$

C
$9V$

D
$0.09V$

Solution

Relative permittivity of the charge $\epsilon_r =10$
Magnitude of charge $q = 10nC = 10^{-8} C$
Potential at a distance $0.1m$ , $V_p = \dfrac{q}{4\pi \epsilon d}$
where $\epsilon = \epsilon_r \epsilon_o$
$\therefore$ $V_p = \dfrac{q}{4\pi \epsilon_o \epsilon_r d}$
Or $V_p = \dfrac{9\times 10^9 \times 10^{-8}}{10 \times 0.1} =90 V$
Question 9
1 / -0

An insulator plate is passed between the plates of a capacitor. Then, current :

A
First flows from A to B and then from B to A

B
First flows from B to A and then from A to B

C
Always flows from B to A

D
Always flows from A to B

Solution
Question 10
1 / -0

Two charges $+6\mu C$ and $-4\mu C$ are placed $15 cm$ apart as shown. At what distances from $A$ to its right, the electrostatic potential is zero?

A
$4, 9, 60$

B
$9, 15, 45$

C
$20, 30, 40$

D
$9, 45$ , infinity

Solution

Let the potential be zero at point $P$ at a distance $x$ from the charge $+6\times { 10 }^{ -6 }C$ at $A$ as shown in above figure. Potential at $P$
$V=\dfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \left[ \dfrac { 6\times { 10 }^{ -6 } }{ x } -\dfrac { \left( -4\times { 10 }^{ -6 } \right) }{ 15-x } \right]$
$0=\dfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \left[ \dfrac { 6\times { 10 }^{ -6 } }{ x } -\dfrac { 4\times { 10 }^{ -6 } }{ 15-x } \right]$
$0=\dfrac { 6\times { 10 }^{ -6 } }{ x } -\dfrac { 4\times { 10 }^{ -6 } }{ 15-x }$
$\dfrac { 6\times { 10 }^{ -6 } }{ x } =\dfrac { 4\times { 10 }^{ -6 } }{ 15-x }$
$\Rightarrow 6\left( 15-x \right) =4x$
$\Rightarrow 90-6x=4x$
$\Rightarrow 10x=90$
$\Rightarrow x=\dfrac { 90 }{ 10 } =9 cm$
The other possibility is that point of zero potential $P$ may lie on $AB$ produced at a distance $x$ from the charge $+6\times { 10 }^{ -6 }C$ at $A$ as shown in the figure.
Potential at $P$
$V=\dfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \left[ \dfrac { 6\times { 10 }^{ -6 } }{ x } +\dfrac { \left( -4\times { 10 }^{ -6 } \right) }{ \left( 15-x \right) } \right]$
$0=\dfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \left[ \dfrac { 6\times { 10 }^{ -6 } }{ x } -\dfrac { 4\times { 10 }^{ -6 } }{ \left( x-15 \right) } \right]$
$0=\dfrac { 6\times { 10 }^{ -6 } }{ x } -\dfrac { 4\times { 10 }^{ -6 } }{ \left( x-15 \right) }$
$\Rightarrow \dfrac { 6\times { 10 }^{ -6 } }{ x } =\dfrac { 4\times { 10 }^{ -6 } }{ x-15 }$
$\Rightarrow \dfrac { 6 }{ x } =\dfrac { 4 }{ x-15 }$
$\Rightarrow 6x-90=4x$
$\Rightarrow 2x=90$
$\Rightarrow x=\dfrac { 90 }{ 2 } =45 cm$

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Electrostatic Potential and Capacitance Test - 47

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Question 1

10017\cfrac { 100 }{ 17 } 17100​

2017V\cfrac { 20 }{ 17 } V1720​V

5017V\cfrac { 50 }{ 17 } V1750​V

10V10V10V

Solution

Question 2

Decreases 222 times

Increases 222 times

Remains same

Becomes zero

Solution

Question 3

(cos⁡αcos⁡β)2\left (\dfrac {\cos \alpha}{\cos \beta}\right )^{2}(cosβcosα​)2

(cos⁡βcos⁡α)2\left (\dfrac {\cos \beta}{\cos \alpha}\right )^{2}(cosαcosβ​)2

(sin⁡αsin⁡β)2\left (\dfrac {\sin \alpha}{\sin \beta}\right )^{2}(sinβsinα​)2

(sin⁡βsin⁡α)2\left (\dfrac {\sin \beta}{\sin \alpha}\right )^{2}(sinαsinβ​)2

Solution

Question 4

(1,n),(2,p+),(3,e−)(1, n), (2, p^{+}), (3, e^{-})(1,n),(2,p+),(3,e−)

(1,p+),(2,e−),(3,n)(1, p^{+}), (2, e^{-}), (3, n)(1,p+),(2,e−),(3,n)

(1,e−),(2,p+),(3,n)(1, e^{-}), (2, p^{+}), (3, n)(1,e−),(2,p+),(3,n)

(1,p+),(2,n),(3,e−)(1, p^{+}), (2, n), (3, e^{-})(1,p+),(2,n),(3,e−)

Solution

Question 5

999

181818

333

272727

Solution

Question 6

74pF\dfrac { 7 }{ 4 } pF47​pF

47pF\dfrac { 4 }{ 7 } pF74​pF

1pF1 pF1pF

2pF2 pF2pF

Solution

Question 7

k+12(k−1)CE\displaystyle\frac{k+1}{2(k-1)}CE2(k−1)k+1​CE

k−12(k+1)CE\displaystyle\frac{k-1}{2(k+1)}CE2(k+1)k−1​CE

k−2k+2CE\displaystyle\frac{k-2}{k+2}CEk+2k−2​CE

k+2k−2CE\displaystyle\frac{k+2}{k-2}CEk−2k+2​CE

Solution

Question 8

900V900V900V

90V90V90V

9V9V9V

0.09V0.09V0.09V

Solution

Question 9

First flows from A to B and then from B to A

First flows from B to A and then from A to B

Always flows from B to A

Always flows from A to B

Solution

Question 10

4,9,604, 9, 604,9,60

9,15,459, 15, 459,15,45

20,30,4020, 30, 4020,30,40

9,459, 459,45, infinity

Solution

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$\cfrac { 100 }{ 17 }$

$\cfrac { 20 }{ 17 } V$

$\cfrac { 50 }{ 17 } V$

$10V$

Decreases $2$ times

Increases $2$ times

$\left (\dfrac {\cos \alpha}{\cos \beta}\right )^{2}$

$\left (\dfrac {\cos \beta}{\cos \alpha}\right )^{2}$

$\left (\dfrac {\sin \alpha}{\sin \beta}\right )^{2}$

$\left (\dfrac {\sin \beta}{\sin \alpha}\right )^{2}$

$(1, n), (2, p^{+}), (3, e^{-})$

$(1, p^{+}), (2, e^{-}), (3, n)$

$(1, e^{-}), (2, p^{+}), (3, n)$

$(1, p^{+}), (2, n), (3, e^{-})$

$9$

$18$

$3$

$27$

$\dfrac { 7 }{ 4 } pF$

$\dfrac { 4 }{ 7 } pF$

$1 pF$

$2 pF$

$\displaystyle\frac{k+1}{2(k-1)}CE$

$\displaystyle\frac{k-1}{2(k+1)}CE$

$\displaystyle\frac{k-2}{k+2}CE$

$\displaystyle\frac{k+2}{k-2}CE$

$900V$

$90V$

$9V$

$0.09V$

$4, 9, 60$

$9, 15, 45$

$20, 30, 40$

$9, 45$ , infinity