Electric Charges and Fields Test

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Electric Charges and Fields Test - 53

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Weekly Quiz Competition

Question 1
1 / -0

the point $$P$$, the flux of the electric field through the closed surface:

A
will remain zero

B
will becomes positive

C
will become negative

D
will become undefined

Solution

We know that the flux of the electric field through the $$Q = EA\cos \theta $$ Here, E is the electric field at point at which electric flux calculate the electric flux at P is undefined because the position of P w.r.t to closed surface is not defined
Question 2
1 / -0

A point charge $$+q$$ is placed at the centre of a cube of side $$L$$. The electric flux emerging from the cube is

A
$$\cfrac{q}{{ \varepsilon }_{ 0 }}$$

B
zero

C
$$\cfrac{6q{L}^{2}}{{ \varepsilon }_{ 0 }}$$

D
$$\cfrac{q}{6{L}^{2}{ \varepsilon }_{ 0 }}$$

Solution

Hint: Electric Flux linked with any surface is the count of Electric Field lines passing through that surface.

Correct Option: A.
Explanation for the correct answer:
$$\bullet$$ It is given that a charge $$+q$$ is placed at the center of the cube of the side $$L$$. Thus, we can say that a charge is placed inside a closed surface, that is the cube.
$$\bullet$$ For closed surface, Gauss Law of Electrostatics states that Electric Flux linked with any closed surface is always $$\dfrac{1}{\epsilon_0}$$ times the total charge enclosed by that surface.
$$\bullet$$ In our question, the cube is a closed surface, and it encloses the charge $$+q$$. Thus, Electric Flux linked with the cube can be given as
$$\phi = \dfrac{1}{\epsilon_0} \times (+q)$$
Or,
$$\phi = \dfrac{+q}{\epsilon_0}$$

Thus, Electric Flux emerging from cube is given as, $$\phi = \dfrac{+q}{\epsilon_0}$$.
Question 3
1 / -0

A point charge of $$40\ stat\ $$ coulomb is placed $$2\ cm$$ in front of an earthed metallic plane plate of large size. Then, the force of attraction on the point charge is

A
$$100\ dynes$$

B
$$160\ dynes$$

C
$$1600\ dynes$$

D
$$400\ dynes$$

Solution

R.E.F image
induced change on the
metallic plate $$ = -40$$ state
$$ \Rightarrow $$ attraction $$ = \dfrac{1}{4\pi w} \times \dfrac{\left | q_{1}q_{2}\right |}{r^{2}}$$
$$ F = 1 \times \dfrac{40 \times 40}{(2)^{2}}$$
$$ \boxed{F = 400 \,dymes }$$
Question 4
1 / -0

A charged particle $$q$$ is projected in an electric field produced by a fixed point charge $$Q$$ as shown in the figure. Select the correct statements:

A
The path taken by $$q$$ is a straight line

B
The path taken by $$q$$ is not a straight line

C
The minimum distance between the two particle is $$\dfrac{\dfrac{qQ}{2\pi\varepsilon_{0}}+\sqrt{\left(\dfrac{qQ}{2\pi\varepsilon_{0}}\right)^{2}+4m^{2}u^{2}a^{2}}}{2mu^{2}}$$

D
Velocity of the particle $$q$$ is changing in both magnitude and direction.
Question 5
1 / -0

Two identical balls having charge $$10mC$$ and $$=1mC$$ are placed at a distance of $$10m$$ from each other, then the force between them will be

A
$$9\times10^2 N$$

B
$$19\times10^2 N$$

C
$$29\times10^2 N$$

D
$$39\times10^2 N$$

Solution

We know that the electrostatic force is given by $$F = \dfrac{q_1\times q_2}{4\pi \epsilon_0\ r^2}$$
F = $$\dfrac{9\times10^9\times10\times10^{-3}\times10^{-3}}{10\times 10}$$
F = $$9\times10^2\ N$$
Question 6
1 / -0

A charged oil drop od mass $$ 2.5\times { 10 }^{ -7 } kg$$ is in space between the two plates, each of area $$ 2\times { 10 }^{ -2 } { m }^{ 2 }$$ of a parallel plate capacitor. When the upper plate has a charge of $$ 5\times { 10 }^{ -7 } C$$ and the lower plate has an equal negative charge then the oil remains stationery. The charge of the oil drop is (take, $$g=10 { m/s }^{ 2 }$$)

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

B
$$ 9\times { 10 }^{ -6 } C$$

C
$$ 8.85\times { 10 }^{ -13 } C$$

D
$$ 1.8\times { 10 }^{ -14 } C$$

Solution

Mass $$=2.5\times { 10 }^{ -7 }kg$$
$$qE=mg$$
or, $$q\times \dfrac { Q }{ A{ \epsilon }_{ 0 } } =mg$$
$$q=\dfrac { mgA{ \epsilon }_{ 0 } }{ Q } $$
$$=\dfrac { 2.5\times { 10 }^{ -2 }\times 2\times { 10 }^{ -2 }\times 8.854\times { 10 }^{ -12 }\times 10 }{ 5\times C7\times 1 } =9\times { 10 }^{ -14 }C$$
Question 7
1 / -0

A point charge $${q_1} = - 5.8\mu C$$ is held stationary at the origin . A second point charge $${q_2} = + 4.3\mu C$$ moves from the point $$(0.26m,0,0)$$ to $$(0.38m,0,0)$$.How much work is done by the electric force on {q_2}.

A
$$2.372J$$

B
$$2.272J$$

C
$$4.783J$$

D
$$1.783J$$

Solution

$${q_1} = 5.8\mu C\,\,\,\,\,\,\,\,{q_2} = 4.3\mu C$$
$${V_1} = \dfrac{{k{q_1}}}{{{r_1}}} = \dfrac{{9 \times {{10}^9} \times 5.8 \times {{10}^{ - 6}}}}{{0.26}}$$
$${V_2} = \dfrac{{k{q_2}}}{{{r_2}}} = \dfrac{{9 \times {{10}^9} \times 5.8 \times {{10}^{ - 6}}}}{{0.38}}$$
$$\Delta = 9 \times {10^9} \times 5.8 \times {10^{ - 6}}\left( {\dfrac{1}{{.26}} - \dfrac{1}{{.38}}} \right)$$
$$ = 2.272J$$
Question 8
1 / -0

The line integral of an electric field along the circumference of a circle of radius r drawn with a point Q at the centre will be _____

A
$$\dfrac{1}{4 \pi^2 to } \dfrac{Q}{r}$$

B
$$\dfrac{Q}{2 \pi \varepsilon_0 r}$$

C
Zero

D
$$2 \pi Q r$$

Solution

$$\begin{array}{l} To\, calculate \\ \oint { \overrightarrow { E } .\overrightarrow { dl } } \\ \oint { \overrightarrow { E } .\overrightarrow { dl } } \, is\, zero\, everywhere\, along\, the\, { { int } }ergral\, will\, be\, zero. \\ \sin ce, \\ \overrightarrow { E } \, and\overrightarrow { dl } \, are\, perpendicular\, to\, each\, other\, \\ \overrightarrow { E } .\overrightarrow { dl } =0 \\ thus \\ \oint { \overrightarrow { E } .\overrightarrow { dl } } =0 \\ Hence,\, the\, correct\, option\, is\, C. \end{array}$$
Question 9
1 / -0

A hemisphere shell is uniformly charged positively. The electric field at a point on a diameter away from the centre (inside the boundary of hemisphere shell) is directed

A
perpendicular to the diameter

B
parallel to the diameter

C
at an angle tilted towards the diameter

D
at an angle tilted away from the diameter

Solution

It has to be perpendicular as when you join the other half to make complete sphere, only this orientation of the electric field will give a vector sum equal to $$0$$, inside the sphere.
Question 10
1 / -0

A square of side 'a' is lying in xy plane that two of its sides are lying on the axis. If $$\vec{E} = E_0 x \hat{k}$$ is applied on the square. The flux passing through the square is

A
$$E_0 a^2$$

B
$$\dfrac{E_0 a^3}{2}$$

C
$$\dfrac{E_0 a^3}{3}$$

D
$$\dfrac{E_0 a^2}{2}$$

Solution

Square side $$=a$$
$$\overline { E } ={ E }_{ 0 }x\hat { k } $$
$$\phi =\overline { E } .\overline { A } $$
$$=\left( { E }_{ 0 }x\hat { k } \right) .\left( { a }^{ 2 } \right) $$
$$={ E }_{ 0 }{ a }^{ 2 }$$