Electric Charges and Fields Test

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Electric Charges and Fields Test - 50

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Question 1
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A charge of magnitude $$q$$ is divided into two parts such that force between resulting two charges is maximum when separated through some distance $$r$$. The division of charges would be

A
$$\large{\frac{3q}{8}}$$, $$\large{\frac{5q}{8}}$$

B
$$\large{\frac{2q}{4}}$$, $$\large{\frac{2q}{4}}$$

C
$$\large{\frac{q}{2}}$$, $$\large{\frac{q}{2}}$$

D
$$\large{\frac{3q}{6}}$$, $$\large{\frac{3q}{6}}$$

Solution

$$\textbf{Step 1: Apply Coulomb's Law } $$
Let $$q$$ is divided into $$q_1$$ and $$(q-q_1)$$
Force between two charges: $$F = \dfrac{1}{4\pi \in_0} \dfrac{q_1(q-q_1)}{r^2} \Rightarrow F = \dfrac{1}{4\pi \in_0} \left( \dfrac{q_1q - q_1^2}{r^2}\right)$$

$$\textbf{Step 2: Maximum force condition} $$
The force, $$F$$, will be maximum if,
$$\dfrac{dF}{dq_1} = 0$$
$$\Rightarrow \dfrac{1}{4\pi \in_0 x^2} \dfrac{d}{dq_1} (q_1q - q_1^2) = 0$$

$$\Rightarrow q - 2q_1 = 0 \Rightarrow q_1 = \dfrac{q}{2}$$

The charges would be $$\dfrac{q}{2}, \dfrac{q}{2}$$.

Hence, option $$(C)$$ is correct.
Question 2
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Five equal charges each of value $$q$$ are placed at the corners of a regular pentagon of side $$a$$. The electric field at the centre of the pentagon is:

A
$$\cfrac { q }{ 4\pi { \varepsilon }_{ 0 }{ r }^{ 2 } } $$

B
$$\cfrac { { q }^{ 2 } }{ 4\pi { \varepsilon }_{ 0 }{ r }^{ 2 } } $$

C
$$\cfrac { 2q }{ 4\pi { \varepsilon }_{ 0 }{ r }^{ 2 } } $$

D
zero

Solution

The point $$O$$, the center of the pentagon is equidistant from all the charges at the end point of pentagon. Thus, due to symmetry, the electric field due to all the charges are cancelled out. As a result electric field at $$O$$ is zero.
Question 3
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A positive charge $$Q$$ is uniformly distributed along a circular ring of radius $$R$$. A small test charge $$q$$ is placed at the centre of the ring as shown in then figure.

A
if $$q> 0$$, and it displaced away from the centre in the plane of the ring, it will be pushed back towards the centre

B
if $$q< 0$$ and is displaced away from the centre in the plane of the ring, it will never return to the centre and will continue moving till it hits the ring

C
if $$q< 0$$ it will perform SHM for small displacement along the axis

D
all of the above

Solution

At the centre of the ring, $$E=0$$ when a positive charge $$q> 0$$ is displaced away from the centre in the plane of the ring, say to the right, force of repulsion on $$q$$, due to charge on right half increases and due to charge on left half decreases. Therefore, charge $$q$$ is pushed back towards the centre. So option (a) is correct.
When charge $$q$$ is negative $$(q< 0)$$, force is of attraction.
Therefore, charge $$q$$ displaced to the right continues moving to the right till it hits the ring. Along the axis of the ring, at a distance $$r$$ from the centre.
$$E=\cfrac { Qr }{ 4\pi { \varepsilon }_{ 0 }{ \left( { r }^{ 2 }+{ a }^{ 2 } \right) }^{ 3/2 } } $$
If charge $$q$$ is negative $$(q< 0)$$, it will perform SHM for small displacement along the axis.
Question 4
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A rectangular surface of sides 10 cm and 15 cm is placed inside a uniform electric field of 25 V/m, such that the surface makes an angle $$30^{\circ}$$ with the direction of electric field. Find the flux of the electric field through the rectangular surface

A
$$0.1675\ \ N m^2/C$$

B
$$0.1875\ \ N m^2/C$$

C
$$Zero$$

D
$$0.1075\ \ N m^2/C$$

Solution

ATQ $$\vec { E } =25V/m$$
$$\vec { A } =10\times { 10 }^{ -2 }\times 15\times { 10 }^{ -2 }=1.5\times { 10 }^{ -2 }{ m }^{ 2 }$$
Now, we know that $$\int { \phi } =\int { Eda } $$
$$=E.A$$
$$=Eacos\theta $$
As the surface make $${ 30 }^{ 0 }$$ but we have to calculate according to area vector which is perpendicular to the surface, $$\therefore \quad \rightarrow \theta ={ 90 }^{ 0 }-{ 30 }^{ 0 }={ 60 }^{ 0 }$$

$$\therefore \quad \phi =Eacos\theta =25\times 1.5\times { 10 }^{ -2 }\times cos{ 60 }^{ 0 }=0.1875\quad { Nm }^{ 2 }/C$$

$$\therefore $$ Option (B) is the correct option.
Question 5
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When a point charge of $$\frac{1}{3}\mu C$$ is placed along the axis of a thin disc of total charge $$\frac{2}{3}\mu C$$ (uniform distribution) and radius 3.95 cm such that distance between the point charge and centre of disc is 1 m, then force experienced by disc is approximately:

A
4mN

B
6mN

C
3mN

D
2mN

Solution

$$\vec E = \dfrac{{k\,qx}}{{{{\left( {{r^2} + {x^2}} \right)}^{\dfrac{3}{2}}}}}$$
$$\vec F = q \cdot \vec E$$
$$ = \dfrac{{kqq \cdot x}}{{{{\left( {{r^2} + {x^2}} \right)}^{\dfrac{3}{2}}}}}$$
$$\vec F = \dfrac{{9 \times {{10}^9} \times \dfrac{{ - 2}}{3} \times {{10}^{ - 6}} \times \dfrac{1}{3} \times {{10}^{ - 6}} \times 1}}{{{{\left[ {{{\left( {0.0395} \right)}^2} + {1^2}} \right]}^{\dfrac{3}{2}}}}}$$
  $$ = \dfrac{{2 \times {{10}^{ - 3}}}}{{{{\left( {0.00156 + 1} \right)}^{\dfrac{3}{2}}}}}$$
  $$ = \dfrac{{2 \times {{10}^{ - 3}}}}{{{{\left( {1.00156} \right)}^{\dfrac{3}{2}}}}}$$
  $$ = \dfrac{{2 \times {{10}^{ - 3}}}}{{1.0023}}$$
  $$ = 1.995 \times {10^{ - 3}}\,N$$
$$\vec F = 2mN$$
Question 6
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In figure two positive charges, $${q}_{2}$$ and $${q}_{3}$$ fixed along the y-axis, exert a net electric force in the $$+x$$ direction on a charge $${q}_{1}$$ fixed along the x-axis. If a positive charge $$Q$$ is added at $$(x,0)$$, the force on $${q}_{1}$$:

A
shall increase along the positive x-axis

B
shall decrease along the positive x-axis

C
shall point along the negative x-axis

D
shall increase but the direction changes because of the intersection of $$Q$$ with $${q}_{1}$$ and $${q}_{3}$$

Solution

Since positive charge $${q}_{2}$$ and $${q}_{3}$$ exert a net force in the $$+X$$ direction on the charge $${Q}_{1}$$ fixed along the X-axis, the charge $${q}_{1}$$ is negative as shown in figure. Obviously, due to addition of positive charge $$Q$$ ad $$(x,0)$$ the force on $$-q$$ shall increase along the positive X-axis.
Question 7
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Two charges $$+20\mu C$$ and $$-20\mu C$$ are placed $$10mm$$ apart. The electric field at point $$P$$, on the axis of the dipole $$10cm$$ away from its centre $$O$$ on the side of the positive charge is

A
$$8.6\times { 10 }^{ 9 }N\quad { C }^{ -1 }$$

B
$$4.1\times { 10 }^{ 6 }N\quad { C }^{ -1 }$$

C
$$3.6\times { 10 }^{ 6 }N\quad { C }^{ -1 }$$

D
$$4.6\times { 10 }^{ 5 }N\quad { C }^{ -1 }$$

Solution

Here $$\quad q=\pm 20\mu C=\pm 20\times { 10 }^{ -6 }C,2a=10mm=10\times { 10 }^{ -3 }m,r=OP=10cm=10\times { 10 }^{ -2 }m$$
$$\left| \vec { p } \right| =q\times 2a=20\times { 10 }^{ -6 }\times 10\times { 10 }^{ -3 }m=2\times { 10 }^{ -7 }m\quad $$
The electric field along $$BP,\vec { E } =\cfrac { 2\vec { p } r }{ 4\pi { \varepsilon }_{ 0 }{ \left( { r }^{ 2 }-{ a }^{ 2 } \right) }^{ 2 } } $$
As $$a<<r$$
$$\vec { E } =\cfrac { 2\left| \vec { p } \right| }{ 4\pi { \varepsilon }_{ 0 }{ r }^{ 3 } } =\cfrac { 2\times 2\times { 10 }^{ -7 }\times 9\times { 10 }^{ 9 } }{ { \left( 10\times { 10 }^{ -2 } \right) }^{ 3 } } =3.6\times { 10 }^{ 6 }N{ C }^{ -1 }\quad $$
Question 8
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A conducting sphere of radius $$10cm$$ has unknown charge. If the electric field at a distance $$20cm$$ from the centre of the sphere is $$1.2\times { 10 }^{ 3 }N { C }^{ -1 }$$ and points radially inwards. The net charge on the sphere is:

A
$$-4.5\times { 10 }^{ -9 }C$$

B
$$4.5\times { 10 }^{ 9 }C$$

C
$$-5.3\times { 10 }^{ -9 }C$$

D
$$5.3\times { 10 }^{ 9 }C$$

Solution

Here distance of the point from the centre of the sphere, $$r=20cm=0.2m$$
Electric field, $$E=-1.2\times { 10 }^{ 3 }N{ C }^{ -1 }\quad $$
$$E=\cfrac { q }{ 4\pi { \varepsilon }_{ 0 }{ r }^{ 2 } } $$
$$\quad \therefore q=\left( 4\pi { \varepsilon }_{ 0 }{ r }^{ 2 } \right) E=\cfrac { { (0.2) }^{ 2 }\times \left( -1.2\times { 10 }^{ 3 } \right) }{ 9\times { 10 }^{ 9 } } =-5.3\times { 10 }^{ -9 }C$$
Question 9
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Two fixed point charges $$ + 16\mu C$$ and $$ - 9\mu C$$ are placed $$8cm$$ apart in air. Where should a third point charge placed so that, the third charge is in equilibrium

A
$$0.24 m$$ from the charge $$ + 16\mu c$$ and within the system of charge

B
$$0.24m$$ from the charge $$ + 16\mu c$$ and outside the system along the straight line joining the two charges

C
$$0.24m$$ from the charge $$ - 9\mu c$$ and within the system of charges

D
$$0.24m$$ from the $$ - 9\mu c$$ and outside the system along the straight lint joining the two charges.

Solution

$${F_{ - q}} = \dfrac{{K\left( { - 9} \right) \times \theta }}{{{{\left( {0.08 + x} \right)}^2}}},\,\,\,\,{F_{ + 16q}} = \dfrac{{K \times 15\mu C}}{{{{\left( x \right)}^2}}}$$
$$\left| {{F_{ - q}}} \right| = \left| {{F_{ + 16q}}} \right| \Rightarrow \,\,\dfrac{9}{{\left( {0.8 + {x^2}} \right)}} = \dfrac{{16}}{{{x^2}}}$$
$${\left( {\dfrac{x}{{0.8 + x}}} \right)^2} = \left( {\dfrac{{16}}{9}} \right) \Rightarrow \,\,\dfrac{x}{{0.8 + x}} = \dfrac{4}{3}$$
$$\boxed{x = 0.24\,m}$$
Question 10
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Two conducting and concentric thin spherical shells or radii $$a$$ and $$b,\,\left( {b > a} \right)$$ have charges $${q_1}$$ and $${q_2}$$ respectively. Now if the inner shell is earthed then the final charge on this shell will be

A
$$\dfrac{{{q_2}{a^2}}}{{{b^2}}}$$

B
$$\dfrac{{ - {q_2}a}}{b}$$

C
$$\dfrac{{\left( {{q_1} + {q_2}} \right)}}{2}$$

D
$$ - \dfrac{{{q_2}b}}{a}$$

Solution

$$\therefore $$ The inner spherical conductor is grounded, $$\Rightarrow $$ potential, $$V=0$$
Also, Potential at inner shell,
$$V=\dfrac { K{ Q }_{ 1 } }{ a } +\dfrac { { KQ }_{ 2 } }{ b } $$
$$\Rightarrow \quad O=\dfrac { K{ Q }_{ 1 } }{ a } +\dfrac { { KQ }_{ 2 } }{ b } $$
$${ Q }_{ 1 }=\dfrac { -{ Q }_{ 2 }a }{ b } $$
$$\therefore $$ Charge on inner shell $${ Q }_{ 1 }=\dfrac { -{ Q }_{ 2 }a }{ b } $$

Option (B) is the correct option.