Electrostatic Potential and Capacitance Test

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Electrostatic Potential and Capacitance Test - 62

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Question 1
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Some charge Q is to be distributed on 3 concentric shell such that surface charge density on each shell is same as shown. Find the electric potential at r(<a) from point O.

A
$$\dfrac { Q(a+b-c) }{ 4\pi { \varepsilon }_{ 0 }({ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }) } $$

B
$$\dfrac { Q(a+b+c) }{ 4\pi { \varepsilon }_{ 0 }({ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }) } $$

C
$$\dfrac { Q(a+b+c) }{ 4\pi { \varepsilon }_{ 0 }({ a }^{ 2 }-{ b }^{ 2 }+{ c }^{ 2 }) } $$

D
$$\dfrac { Q(a+b-c) }{ 4\pi { \varepsilon }_{ 0 }({ a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 }) } $$

Solution

$$r<a$$
from point $$0$$.
$$\oint { \overline { E } } .d\overline { A } =\dfrac { { Q }_{ inside } }{ { \epsilon }_{ 0 } } $$
$${ \epsilon }_{ 0 }=8.854\times { 10 }^{ -12 }\dfrac { { e }^{ 2 } }{ { Nm }^{ 2 } } $$
Now, $$E\times 4\pi { R }_{ 1 }^{ 2 }=\dfrac { -5\mu C }{ { \epsilon }_{ 0 } } =\dfrac { { Q }_{ inside } }{ { \epsilon }_{ 0 } } $$
$$E=\dfrac { { Q }_{ inside } }{ 4\pi { \epsilon }_{ 0 }{ R }_{ 1 }^{ 2 } } $$
$${ E }_{ 1 }=\dfrac { K{ Q }_{ incide } }{ { R }_{ 1 }^{ 2 } } $$
$${ E }_{ 2 }A=\dfrac { { Q }_{ inside } }{ { \epsilon }_{ 0 } } $$
$${ F }_{ y }4\pi { R }_{ z }^{ 2 }=\dfrac { Q }{ { \epsilon }_{ 0 } } $$
Now, $$\dfrac { Q\left( a+b+c \right) }{ 4\pi { \epsilon }_{ 0 }\left( { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } \right) } $$
Question 2
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A capacitor of capacitance C = 4 $$\mu F$$ is connected as shown in figure. If internal resistance of the cell is 0.5 . The charge on the capacitor plates is

A
zero

B
8 $$\mu \mathrm { C }$$

C
4 $$\mu \mathrm { C }$$

D
6 $$\mu \mathrm { C }$$

Solution

At steady state the capacitor is open circuited so no current flows through the $$10$$ $$ohm$$ resistor$$.$$
Current flowing through $$2$$ $$ohm$$ resistor is
$$i = \frac{{2.5}}{{2 + 0.5}} = 2A$$
Voltage across capacitor terminal$$:$$ $$2.5 - 0.5 \times 1 = 2V$$
Charge in capacitor plates $$4 \times {10^{ - 6}} \times 2 = 8\,\mu C$$
Hence,
option $$(B)$$ is correct answer.
Question 3
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The radius of the gold nucleus is $$6.6 \times {10^{ - 15}}m$$ and the atomic number is $$79$$. The electric potential at the surface of the gold nucleus is :

A
$$1.7 \times {10^7}V$$

B
$$7.1 \times {10^7}V$$

C
$$1.7 \times {10^9}V$$

D
$$7.1 \times {10^9}V$$

Solution
Question 4
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Two plates of equal area A , one a square of side $$2L \times 2L $$ and other a rectangle os sides $$ 4L \times L $$ are used to construct a parallel plate capacitor with plates separation . d . if C is the capacitance , then C=

A
$$ \frac { \varepsilon _{ 0 }A }{ 2d } $$

B
$$ \frac { \varepsilon _{ 0 }A }{ d } $$

C
$$ \frac {2 \varepsilon _{ 0 }A }{ d } $$

D
none of these
Question 5
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Four capacitors are joined as shown in the adjoining figure. The capacitance of each is $$ 8\mu F $$. The equivalent capacitance between the points A and B is

A
$$ 32\mu F $$

B
$$ 2\mu F$$

C
$$ 8\mu F $$

D
$$ 16\mu F $$

Solution
Question 6
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A combination arrangement of the capacitors is shown in the figure.
(i) $$ C_1=3 \mu F , C_2=6 \mu F and C_3 = 2 \mu F $$ then equivalent capacitance between 'a'and 'b' is

A
$$ 4 \mu F $$

B
$$ 6 \mu F $$

C
$$ 1 \mu F $$

D
$$ 2 \mu F $$

Solution
Question 7
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In the given figure find $$V_{A} - V_{B}$$.

A
$$5V$$

B
$$-5V$$

C
$$3V$$

D
$$-3V$$
Question 8
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An infinite number of charges 'q' each are placed along the x - axis at x =1, x=4 ,x=8 and so on. If the distance are in meters calculate the electric potential at x=0

A
$$\frac{{3q}}{{8\pi {E_0}}}$$

B
$$\frac { q } { 2 \pi \epsilon _ { 0 } }$$

C
$$\frac { 2 q } { \pi \epsilon _ { 0 } }$$

D
$$\frac { 4 q } { \pi \epsilon _ { 0 } }$$

Solution

$$\begin{array}{l} { V_{ 0 } }=\frac { { kq } }{ 1 } +\frac { { kq } }{ 4 } +\frac { { kq } }{ 8 } +....................... \\ =kq\left( { 1+\frac { 1 }{ { { 2^{ 2 } } } } +\frac { 1 }{ { { 2^{ 3 } } } } +.......... } \right) \\ =kq\left( { 1+\frac { { \frac { 1 }{ 4 } } }{ { 1-\frac { 1 }{ 2 } } } } \right) \\ =kq\left( { 1+\frac { 2 }{ 4 } } \right) \\ =\frac { { 3kq } }{ 2 } =\frac { { 3q } }{ { 8\pi { E_{ 0 } } } } \end{array}$$
Hence, Option $$A$$ is correct.
Question 9
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In the situation shown in figure, what should be the relation between Q and q so that electric potential at centre of the square is zero:

A
$$Q=q$$

B
$$Q=3q$$

C
$$Q=2q$$

D
$$Q=-3q$$

Solution

$$Q=$$ Sum motion of three side total charges
$$=q+q+q$$
$$=3q$$
Question 10
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In the shown network chages in capacitor are same, then $$ \frac {C_1}{C_2} is $$

A
$$ \frac {5}{3} $$

B
1 : 1

C
1 : 3

D
1 : 5