Electric Charges and Fields Test

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Electric Charges and Fields Test - 58

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Question 1
1 / -0

A charge of magnitude $$3e$$ and mass $$2m$$ is moving in an electric field $$E$$. The acceleration imparted to the charge is

A
$$ \dfrac { 2Ee }{ 3m } $$

B
$$ \dfrac { 3Ee }{ 2m } $$

C
$$ \dfrac { 2m }{ 3Ee } $$

D
$$ \dfrac { 3m }{ 2Ee } $$

Solution

As we know,
Acceleration $$a=\dfrac{QE}{m}=\dfrac{(3e)E}{2m}$$
Question 2
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A charge $$Q$$ is placed at the center of a square. If electric field intensity due to the charge at the corners of the square is $$ E_1 $$ and the intensity at the mid point of the side square is $$ E_2 $$ then the ratio of $$ \dfrac { E_ 1 }{ E_ 2 }$$ will be

A
$$ \dfrac { 1 }{ 2\sqrt { 2 } } $$

B
$$ \sqrt { 2 } $$

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

D
$$2$$

Solution

$$\textbf{Step 1: Calculate all the distances[Ref. fig]} $$
Distance of charge $$Q$$ from corner $$D$$ :
$$r_{1} = OD = \dfrac{BD}{2} = \dfrac{\sqrt{a^2 + a^2}}{2} = \dfrac{a}{\sqrt{2}} $$

Distance of charge $$Q$$ from midpoint $$E$$ of the side $$AD$$:
$$r_{2} = OE = \dfrac{a}{2} $$

$$\textbf{Step 2: Electric field at corner D} $$

$$E_{1} = \dfrac{1}{4 \pi \epsilon_0} \dfrac{Q}{r_{1}^{2}} = \dfrac{1}{4 \pi \epsilon_0} \dfrac{Q}{(a /\sqrt{2})^2} = \dfrac{2Q}{(4 \pi \epsilon_0) a^2} $$ $$....(1)$$

$$\textbf{Step 3: Electric field at midpoint E of side AD}$$

$$E_2 = \dfrac{1}{4 \pi \epsilon_0} \dfrac{Q}{r_2^2} = \dfrac{1}{4 \pi \epsilon_0} \dfrac{Q}{(a / 2)^2} $$ $$ = \dfrac{4Q}{(4 \pi \epsilon_0) a^2} $$ $$....(2)$$

$$\textbf{Step 4: Ratio of}\, E_{1}\ \&\ E_{2}$$

Dividing eq $$(1)$$ and $$(2)$$

$$\dfrac{E_1}{E_2} = \dfrac{1}{2} $$

Hence, Option (C) is correct.
Question 3
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A hemispherical surface of radius R is kept in a uniform electric field E as shown in figure. The flux through the curved surface is:

A
$$E2\pi R^2$$

B
$$E\pi R^2$$

C
$$E4\pi R^2$$

D
$$Zero$$

Solution

Flux passing through the curved surface is equal to the flux passing through the flat surface.
Since flux through the flat surface $$\phi_{flat} = E\times Area = E\times \pi R^2$$
Flux through the curved surface $$\phi_{curved} = \phi_{flat} = E\times \pi R^2$$
Correct answer is option B.
Question 4
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A point charge $$Q$$ is placed at the centre of a circular wire of radius $$R$$ having charge $$q$$. The force of electrostatic interaction between point charge and the wire is:

A
$$\cfrac{qQ}{4\pi\varepsilon_oR^2}$$

B
$$\cfrac{q}{4\pi\varepsilon_oR}$$

C
Zero

D
none of these

Solution

A point charge $$=Q$$
radius $$=R$$
$$F=\dfrac { 1 }{ 4\pi { \epsilon }_{ 0 } } \dfrac { Q\times X }{ { R }^{ 2 } } $$
There is only one point charge. But according to Coulomb's law interaction force create with two point charges hence none of this is correct.
Question 5
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Two charges Q and -2Q are placed at some distance. the locus of points in the plane of the charges where the potential is zero will be

A
Straight line

B
Circle

C
Parabola

D
ellipse

Solution

Two charges $$Q$$, $$-2Q$$
some distance
Potential $$=0$$
hence,
$$\dfrac { 1 }{ 4\pi { \epsilon }_{ 0 } } \times \dfrac { Q\times \left( -2Q \right) }{ R } =0$$
When we use it the parabolic condition then,
$${ y }^{ 2 }=4ax\quad \longrightarrow \left( 1 \right) $$
Now,
$$\dfrac { 1 }{ 4\pi { \epsilon }_{ 0 } } \times \dfrac { Q\times 2Q }{ R } =0$$
$$\dfrac { { 2Q }^{ 2 } }{ R } =0\quad \longrightarrow \left( 2 \right) $$
Hence, equating $$(1)$$ and $$(2)$$ and we get, the system is getting parabola.
Question 6
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The electric potential is directly proportional to from the region. The intensity of electric field

A
Is proportional to $$r^{2}$$

B
Is changing with uniform rate

C
Is uniform

D
Is non-uniform

Solution
Question 7
1 / -0

The tangent at any point of field line gives the direction of

A
Electric field at that point

B
Electric force on positive charge at that point

C
Both $$(1)$$ & $$(2)$$

D
Rotation of charge

Solution
Question 8
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Under the influence of the Coulomb field of charge $$+Q$$, a charge $$-q$$ is moving around it in an elliptical orbit. Find out the correct statements.

A
The angular momentum of the charge $$-q$$ is constant.

B
The linear momentum of the charge $$-q$$ is constant.

C
The angular velocity of the charge $$-q$$ is constant.

D
The linear speed of the charge $$-q$$ is constant.

Solution

Coulomb field of charge $$=+Q$$
a charge $$=-q$$
The angular momentum of the charge $$-q$$ is constant.
angular momentum of the charge $$=(-q)$$ $$K$$.
$$F=\dfrac { kq\left( -Q \right) }{ { r }^{ 2 } } $$
angular momentum $$=K(mV)$$
Question 9
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A charge Q is placed a the corner of a cube. The electric flux through all the six faces of the tube is

A
$$\frac{Q}{{{\varepsilon _0}}}$$

B
$$\frac{Q}{{{6\varepsilon _0}}}$$

C
$$\frac{Q}{{{8\varepsilon _0}}}$$

D
$$\frac{Q}{{{3\varepsilon _0}}}$$

Solution

The charge is kept at a corner of a cube, so it can be visualized as being at the center of a cube of a double its side i. e., comprising of 8 such cubes kept besides and top of each other.
The flux to each small cube is $$\dfrac{Q}{8 \varepsilon_0}$$.
The correct option is C.
Question 10
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A hollow cylinder has a charge q coulomb within it. If f the electric flux in units of volmeter associated with the curved surface B , the flx linked with the plane surface A in units of V-m will be :

A
$$\dfrac { q }{ 2{ \varepsilon }_{ 0 } } $$

B
$$\dfrac { \Phi }{ 3 } $$

C
$$\dfrac {q}{\varepsilon _0}$$

D
$$\dfrac {4q}{\varepsilon_0}$$

Solution

Charge $$=qC$$
associated with the curved surface $$=B$$
Now, using Gauss law,
$$\phi =$$ the flux linked with the plane surface $$A$$ in units of $$V-m$$
$$=\dfrac { q }{ { \epsilon }_{ 0 } } $$