Electrostatic Potential and Capacitance Test

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Electrostatic Potential and Capacitance Test - 51

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Question 1
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The ratio of momentum of an electron and an alpha particle which are accelerated from rest by potential difference of 100 V is:

A
$$\sqrt{\dfrac{m_{\alpha}}{m_e}}$$

B
$$\sqrt{\dfrac{m_e}{m_{\alpha}}}$$

C
$$\dfrac{2m_e}{m_{\alpha}}$$

D
$$\sqrt{\dfrac{m_e}{2m_{\alpha}}}$$

Solution

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
$$P_2=\sqrt{2qVm_2}=\sqrt{2\times 2eVm_a}$$
$$\therefore \cfrac{P_1}{P_2}=\sqrt{\cfrac{2eVm_e}{4eVm_a}}$$
$$\therefore \cfrac{P_1}{P_2}=\sqrt{\cfrac{m_e}{2m_a}}$$
Question 2
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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.

A
r

B
$$2r$$

C
$$r/2$$

D
$$r/4$$

Solution

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, $$\displaystyle r'=\frac{1}{4\pi \varepsilon_0}\frac{Qq}{m(2v)^2}$$
$$\Rightarrow r'=\displaystyle \frac{1}{4}\left(\displaystyle\frac{1}{4\pi \varepsilon_0}\cdot \frac{Qq}{mv^2}\right)\Rightarrow r'=r/4$$.
Question 3
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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

A
$$\cfrac { \pi { \varepsilon }_{ 0 }\lambda A }{ 2d } $$

B
$$\cfrac { \pi { \varepsilon }_{ 0 }\lambda A }{ d } $$

C
$$\cfrac { 2\pi { \varepsilon }_{ 0 }\lambda A }{ d } $$

D
None

Solution

$$E=\cfrac{\sigma}{K\epsilon_o}$$
$$\cfrac{dv}{dy}=\cfrac{\sigma}{K\epsilon_o}$$
$$\int _{ 0 }^{ v } dv=\cfrac { \sigma }{ \lambda \epsilon _{ o } } \int _{ 0 }^{ d } \cos \left( \cfrac { \pi y }{ 2d } \right) dyV\\ =\cfrac { \sigma }{ \lambda \epsilon _{ o } } \times \cfrac { 2d }{ \pi } \left[ \sin \cfrac { \pi y }{ 2d } \right] _{ 0 }^{ d }\\ =\cfrac { \sigma }{ \lambda \epsilon _{ o } } \times \cfrac { 2d }{ \pi } \\ C=\cfrac{ Q}{V}=\cfrac{A\lambda\epsilon_o\pi}{2d}$$
Question 4
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A capacitor is charged by a battery and then the battery is disconnected. A dielectric slab is introduced between the plates. The result is

A
P.d between the plates increases, charge on the plate decreases

B
P.d between the plates decreases, charge remains same

C
P.d increases, charge remain constant

D
P.d decreases, charge increases

Solution

As dielectric slab does not affect charges:
$$V=\cfrac{Q}{C}=\cfrac{Qd}{K\epsilon_oA}$$
Question 5
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The equivalent capacitance between point $$a$$ and $$b$$ in the combination of capacity figure is

A
$$15\mu F$$

B
$$10\mu F$$

C
$$11.2\mu F$$

D
$$7.4\mu F$$

Solution

FOR IMAGE 01: The series combination are:
$$\left[C_s=\cfrac{2\times 3}{2+3}=\cfrac{6}{5}\mu F\right]$$
FOR IMAGE 02: The parallel Combination are:
$$C_{eq}=4+\cfrac{6}{5}+5$$
$$=\cfrac{56}{5}=11.2\mu F$$
Question 6
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The equivalent capacitance for the network shown in the figure is:

A
$$\dfrac { 300 }{ 4 }\ pF$$

B
$$\dfrac { 1000 }{ 4 }\ pF$$

C
$$\dfrac { 1800 }{ 7 }\ pF$$

D
$$\dfrac { 1300 }{ 3 }\ pF$$

Solution

Capacitance of $${C}_{1}={C}_{4}=100pF$$
Capacitance of $${C}_{2}={C}_{3}=400pF$$
Supply voltage, $$V=400V$$
Capacitors $${C}_{2}$$ and $${C}_{3}$$ are connected in series
Equivalent capacitance
$$C'=\cfrac { 1 }{ 400 } +\cfrac { 1 }{ 400 } =\cfrac { 2 }{ 400 } $$
or $$C'=200pF$$
Capacitors $${C}_{1}$$ $$C'$$ are in parallel
Their equivalent capacitance
$$C''=C'+{C}_{1}=200+100=300pF$$
Capacitors $$C''$$ and $${C}_{4}$$ are connected in series
Equivalent capacitance $$\cfrac { 1 }{ { C }_{ eq } } =\cfrac { 1 }{ C'' } +\cfrac { 1 }{ { C }_{ 4 } } =\cfrac { 1 }{ 300 } +\cfrac { 1 }{ 100 } $$
$$\cfrac { 1 }{ { C }_{ eq } } =\cfrac { 4 }{ 300 } $$
$$\therefore { C }_{ eq }=\cfrac { 300 }{ 4 } pF$$
Question 7
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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

A
$$ \dfrac { { C }_{ 1 }V }{ { C }_{ 1 }+{ C }_{ 2 } }$$

B
$$ \dfrac { { C }_{ 2 }V }{ { C }_{ 1 }+{ C }_{ 2 } }$$

C
$$ 1+\dfrac { { C }_{ 2 } }{ { C }_{ 1 } }$$

D
$$ 1-\dfrac { { C }_{ 2 } }{ { C }_{ 1 } }$$

Solution
Question 8
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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

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

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

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

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

Solution

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))
Question 9
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Two capacitors of 2 $$\mu F$$ and 4 $$\mu F$$ are connected in parallel. A third capacitor of 6 $$\mu F$$ is connected in series. The combination is connected across a 12 V battery. The voltage across 2 $$\mu F$$ capacitor is:

A
2 V

B
8 V

C
6 V

D
1 V

Solution

$${ C }_{ P }=2+4=6\mu F$$
$$\cfrac { 1 }{ C } =\cfrac { 1 }{ 6 } +\cfrac { 1 }{ 6 } =\cfrac { 2 }{ 6 } =\cfrac { 1 }{ 3 } \quad or\quad C=3\mu F$$
Total charge
$$Q=CV=3\times 12=36\mu C\quad \quad $$
voltage across $$6\mu F$$ capacitor $$\quad =\cfrac { 36\mu C }{ 6\mu F } =6V$$
$$\therefore$$ Voltage across each of $$2\mu F$$ and $$4\mu F$$ capacitors
$$=12V-6V$$
$$V=6V$$
Question 10
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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

A
2 $$\mu C$$

B
3 $$\mu C$$

C
5 $$\mu C$$

D
10 $$\mu C$$

Solution

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