Electric Charges and Fields Test

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Electric Charges and Fields Test - 70

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Question 1
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Two non-conducting plates $$A$$ and $$B$$ of radii $$2R$$ and $$4R$$ respectively are kept at distances $$x$$ and $$2x$$ from the point charge $$q$$. A surface cutout of a non conducting shell $$C$$ is kept such that its centre coincides with the point charge. Each plate and the spherical surface carries a surface charge density $$\sigma$$. If $$\phi_{1}$$ is flux through surface of $$(B)$$ due to electric field of $$(A)$$ and $$\phi_{2}$$ be the flux through $$(A)$$ due to electric field of $$(B)$$ then:

A
$$\phi_{1} = \phi_{2}$$

B
$$\phi_{1} > \phi_{2}$$

C
$$\phi_{1} < \phi_{2}$$

D
It depend and on $$x$$ and $$R$$

Solution

From the defination of flux,
$$\phi=E\cdot dA$$
or, $$\phi= E(A)$$
where, $$A$$ is the area of surface.
Now, according to question,
$$\phi _1:$$ Flux through $$B$$ due to $$A$$
So, $$E$$ due to $$A=\cfrac{\sigma}{2\varepsilon_o}$$ (Non- conducting charge sheet)
$$\Rightarrow \phi_1=E$$ (Area of $$B$$)
$$\Rightarrow \phi_1=\cfrac{\sigma}{\varepsilon_o}(\pi 4R^2)=\cfrac{2\sigma \pi R^2}{\varepsilon_o}$$
Similarly for $$\phi_2$$: Flux through $$A$$ due to $$B$$,
$$E$$ due to $$B=\cfrac{\sigma}{2\varepsilon_o}$$
$$\Rightarrow \phi_2=E$$ (Area of $$A$$)
$$\Rightarrow \phi_2=\cfrac{\sigma}{2\varepsilon_o}\left[16R^2\pi\right]=\cfrac{8\sigma \pi R^2}{\varepsilon_2}$$
So, $$\phi_2>\phi_1$$
Question 2
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A point charge $$Q(C)$$ is placed at the origin. Find the electric flux of which an area $$4\pi\ m^2$$ on a concentric spherical shell of radius $$R$$

A
$$\large{\frac{Q}{R^2 \epsilon_0}}$$

B
$$\large{\frac{Q}{ \epsilon_0}}$$

C
$$\large{\frac{Q}{4R^2 \epsilon_0}}$$

D
none of above

Solution

From Gauss law,
$$\phi =\dfrac { { Q }_{ inc } }{ { \varepsilon }_{ 0 } } $$
For a spherical shell
$$\rightarrow $$ flux through an area of $${ 4\pi r }^{ 2 }$$
$${ m }^{ 2 }=\dfrac { Q }{ { \varepsilon }_{ 0 } } $$
$$\rightarrow $$ Flux through 1 $${ m }^{ 2 }$$ area $$=\dfrac { Q }{ 4\pi { R }^{ 2 }{ \varepsilon }_{ 0 } } $$
Also flux through $$4\pi { m }^{ 2 }=\dfrac { 4\pi Q }{ 4\pi { R }^{ 2 }{ \varepsilon }_{ 0 } } =\dfrac { Q }{ { R }^{ 2 }{ \varepsilon }_{ 0 } } $$
$$\therefore $$ we can conclude that flux through an area of $$4\pi { m }^{ 2 }$$ on a shell is given by $$=\dfrac { Q }{ { R }^{ 2 }{ \varepsilon }_{ 0 } } $$

$$\therefore $$ Option (A) is the correct answer.
Question 3
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Force$$ = F,$$ When a dielectric rod is placed between charges as shown. new force

A
$$ > F$$

B
$$ < F$$

C
$$ = F$$

D
Nothing can be said

Solution

$${F_{new}} = \dfrac{1}{{4\pi k{\varepsilon _0}}}\dfrac{{q\left( { - q} \right)}}{{{r^2}}}$$
$${F_{new}} = \dfrac{{{F_{old}}}}{{{\varepsilon _0}}}$$
So, $$\boxed{{F_{new}} = {F_{old}}}$$
Question 4
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Two similar very small conducting spheres having charges $$40\,\mu C$$ and $$ - 20\,\mu C$$ are some distance apart. Now they are touched and kept at the same distance. The ratio of the initial to the final force between them is:

A
$$8:1$$

B
$$4:1$$

C
$$1:8$$

D
$$1:1$$

Solution

Initial force $$=\dfrac { K\left( 40\times { 10 }^{ -6 } \right) \left( -20\times { 10 }^{ -6 } \right) }{ { r }^{ 2 } } =\dfrac { 8K\times { 10 }^{ -10 } }{ { r }^{ 2 } } N$$
when they are touched the charge redistrubuted among them & final charge on both of them is average of them.
$$\Rightarrow \quad $$ Charge on one sphere $$=\left( \dfrac { 40-20 }{ 2 } \right) =10\mu c$$
$${ 2 }^{ nd }$$ sphere $$=10\mu c$$
Final force $$=\dfrac { \left( K \right) \left( 10\times { 10 }^{ -6 } \right) \left( 10\times { 10 }^{ -6 } \right) }{ { r }^{ 2 } } =\dfrac { \left( K \right) \left( { 10 }^{ -10 } \right) }{ { r }^{ 2 } } $$
$$\Rightarrow \quad $$ Initial force : Final force $$=8:1$$
Question 5
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Three point charges $${q_1},{q_2}$$ and $${q_3}$$ are taken such that when $${q_1}$$ and $${q_2}$$ are placed close together to form a single point charge, the force on $${q_3}$$ at distance $$L$$ from this combination is a repulsion of $$2$$ unit magnitude. When $${q_2}$$ and $${q_3}$$ are combined, the force on $${q_1}$$, from same distance $$L$$ is of magnitude $$4$$ unit and is attractive. Also $${q_3}$$ and $${q_1}$$ combining exert $$6$$ a force of magnitude $$18$$ unit attractive on $${q_2}$$ from same distance $$L$$, Algebraic ratio of charges $${q_1}, {q_2}$$ and $${q_3}$$ is:

A
$$1 : 2 : 3$$

B
$$2 : -3 : 4$$

C
$$4 : -3 : 1$$

D
$$4 : -3 : 2$$

Solution

$$\dfrac{{{q_3}\left( {{q_1} + {q_2}} \right)}}{{{L^2}}} = 2$$
$$\dfrac{{{q_1}\left( {{q_2} + {q_3}} \right)}}{L} = - 4$$
$$\dfrac{{{q_2}\left( {{q_1} + {q_3}} \right)}}{{{L^2}}} = 18$$
$$\dfrac{{{K_1}\left( {{K_1} + 1} \right)}}{{{K_1}\left( {1 + {K_2}} \right)}} = \dfrac{1}{9}$$
$$\dfrac{{{K_1} + {K_2}}}{{{K_2}\left( {{K_1} + 1} \right)}} = - 2$$
$${K_2} = \left( {\dfrac{{{K_1}}}{{8{K_1} + 9}}} \right) = 2$$
$${q_1}:{q_2}:{q_3} = 1:\dfrac{{ - 6}}{5}:2$$
Question 6
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The flux of the electric field due to charges distributed in a sphere of radius $$5$$ cm is $$10$$ Vm. What will be the electric flux, through a concentric sphere of radius $$10$$ cm ?

A
$$20\ Vm$$

B
$$30 \ Vm$$

C
$$5\ Vm$$

D
$$10 \ Vm$$

Solution

Given :- $${ \phi }_{ A }=10Vm$$
$$r=5cm=5\times { 10 }^{ -2 }m$$
Now, from Gauss law,
$$\phi =\dfrac { { Q }_{ inc } }{ { \varepsilon }_{ 0 } } $$
$$\therefore $$ we can conclude from Gauss law that flux is a quantity which depend on the total charge inside the gaussion surface.

Now, Initially charge contained is same as finally,
$$\therefore $$ flux will also be some,
$$\Rightarrow { \phi }_{ B }=10Vm$$ through a surface of $$10cm$$

$$\therefore $$ Option (D) is the correct answer.
Question 7
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The electric flux emerging through the closed surface $${S_1}$$ shown in figure which intersects the spherical conductor $$S,$$ due to the presence of a positive charge very near to conductor is:

A
Positive

B
Negative

C
Zero

D
None of the above

Solution

As you see, due to $$+q$$ charge a $$-ve$$ charge will induce on the surface of conductor which is near to the charge.
But there will be a positive charge also induced on the other side and our gaussion surface will thus have positive charge enclosed in it.
$$\Rightarrow \quad \phi =\dfrac { +q }{ { \varepsilon }_{ 0 } } $$ {positive flux}

$$\therefore $$ Option A is the correct answer.
Question 8
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A number of spherical conductors of different radii are changed to same potential. The surface charge density of each conductor is related with its radius as

A
$$\sigma \propto \dfrac{1}{R^2}$$

B
$$\sigma \propto \dfrac{1}{R}$$

C
$$\sigma \propto R$$

D
None of these

Solution

$$\sigma=\cfrac{q}{A}\,\,\,\,\sigma=\cfrac{q}{4\pi r^{2}}$$(as sphere).
$$v=\cfrac{q}{4\pi\epsilon_{0}r}=\cfrac{\sigma r}{\epsilon_{0}}$$
$$\therefore \sigma=\cfrac{v\epsilon_{0}}{r}$$(As $$v$$ is constant).
$$\therefore \sigma\propto\cfrac{1}{r}$$
Question 9
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Three infinitely charged sheets are kept parallel to $$x-y$$ plane having charge densities as shown. Then the value of electric field at $$'P'$$ is:

A
$$\dfrac { -4\sigma }{ { \in }_{ 0 } } \hat { k }$$

B
$$\dfrac { 4\sigma }{ { \in }_{ 0 } } \hat { k }$$

C
$$\dfrac { -2\sigma }{ { \in }_{ 0 } } \hat { k }$$

D
$$\dfrac { 2\sigma }{ { \in }_{ 0 } } \hat { k }$$

Solution

$$E$$ at $$P$$
$$=E_{dw\quad to Z=0}+E_{Z=2a}+E_{Z=3a}$$
$$=-\cfrac{\sigma}{2\epsilon_o}-\cfrac{2 \sigma}{2\epsilon_o}-\cfrac{1 \sigma}{2\epsilon_o}$$
$$=-\cfrac{4\sigma}{2\epsilon_o}\hat K$$
$$=-\cfrac{-2\sigma}{\epsilon_o}\hat K$$
Question 10
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Two point charges placed at a distance r in air expert a force F on each other. The value of distance R at which they experience force $$4F$$ when placed in a medium of dielectric constant $$K = 16$$ is:

A
$$r$$

B
$$\dfrac{r}{4}$$

C
$$\dfrac{r}{8}$$

D
$$2r$$

Solution

Two points charge placed at a distance $$=r$$
force $$=F$$
value of distance $$=R$$
$$F=\dfrac { 1 }{ 4\pi { \epsilon }_{ 0 } } \dfrac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } $$
$$4F=\dfrac { 1 }{ 4\pi { \epsilon }_{ 0 }\times K } \dfrac { { q }_{ 1 }{ q }_{ 2 } }{ { R }^{ 2 } } $$
or,  $$F=\dfrac { 1 }{ 4\pi { \epsilon }_{ 0 }\times 4\times 16 } \times \dfrac { { q }_{ 1 }{ q }_{ 2 } }{ { R }^{ 2 } } $$
or, $$\dfrac { 1 }{ 4\pi { \epsilon }_{ 0 } } \dfrac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } $$ $$=\dfrac { 1 }{ 64 } \dfrac { 1 }{ 4\pi { \epsilon }_{ 0 } } \dfrac { { q }_{ 1 }{ q }_{ 2 } }{ { R }^{ 2 } } $$
$$=\dfrac { 1 }{ 64 } \left( \dfrac { 1 }{ 4\pi { \epsilon }_{ 0 } } \dfrac { { q }_{ 1 }{ q }_{ 2 } }{ { R }^{ 2 } } \right) $$
or,  $$\dfrac { 1 }{ { r }^{ 2 } } =\dfrac { 1 }{ 64 } \times \left( \dfrac { 1 }{ 4\pi { \epsilon }_{ 0 } } \dfrac { { q }_{ 1 }{ q }_{ 2 } }{ { R }^{ 2 } } \right) $$
or,  $$\dfrac { 1 }{ { r }^{ 2 } } =\dfrac { 1 }{ 64 } \times \dfrac { 1 }{ { R }^{ 2 } } $$
$$\dfrac { 1 }{ r } =\dfrac { 1 }{ 8 } \dfrac { 1 }{ R } $$
or,  $$\boxed { R=\dfrac { r }{ 8 } } $$