Electric Charges and Fields Test

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Electric Charges and Fields Test - 60

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following is false for electric lines of force ?

A
They always start from positive charges and terminate on negative charges

B
They are always perpendicular to the surface of a charged conductors

C
They always form closed loops

D
They are parallel and equally spaced in a region of uniform electric field

Solution
Question 2
1 / -0

There are two charges +2 $$\mu C$$ and -3$$\mu C$$ .The ratio of forces acting on them will be

A
2 : 3

B
1 : 1

C
3 : 2

D
4 : 9

Solution
Question 3
1 / -0

If the surface normal vector $$ \hat{n} $$ makes an angle $$ \theta $$ withthe electric field $$ \vec{E} $$ , then the electric flux through asurface of area $$ d S $$ is given by

A
$$ \phi=E d S \sin \theta $$

B
$$ \phi=E d S \cos \theta $$

C
$$ \phi=d S \cos \theta $$

D
$$ \phi=E \cos \theta $$

Solution
Question 4
1 / -0

An electric field exists in the space between two charged metal plates.
which graph shows the variation of electric field strenth E with distance d from X along the line XY ?

A

B

C

D

Solution

Strength $$=E$$
distance $$=d$$
$$X$$ along the line $$XY$$
$$E$$ $$\alpha$$ $$\dfrac { 1 }{ { d }^{ 2 } } $$ in this context, the graph is the exact like this.
Question 5
1 / -0

A charge q is placed at the center of the line joining two equal charges Q . The system of the three charges will be equilibrium, If q is equal to :-

A
$$-\dfrac { Q }{ 2 } $$

B
$$-\dfrac { Q }{ 4 } $$

C
$$+\dfrac { Q }{ 4 } $$

D
$$+\dfrac { Q }{ 2 } $$

Solution

The correct option should be B
Question 6
1 / -0

A problem of practical interest is to make a beam of electrons turn at $$90^{\circ}$$ corner. This can be done with the parallel plates shown in figure. An electron with kinetic energy $$8.0\times 10^{-17} J$$ enters through a small hole in the bottom plate. The strength of electric field that is needed if the electron is to emerge from an exit hole $$1.0\ cm$$ away from the entrance hole, traveling at right angles to its original direction is _______ $$\times 10^{5} N/C$$.

A
$$3$$

B
$$1$$

C
$$6$$

D
$$2.5$$
Question 7
1 / -0

The flux through the hemispherical surface in the figure shown below is

A
$$\pi R^2E$$

B
$$2\pi R^2E$$

C
$$\pi RE$$

D
Zero

Solution

$$\textbf{Hint}$$: Flux can be calculated by the product of electrostatic field intensity and the area.
$$\textbf{Step 1}$$:
Here there is no charge inside the hemisphere. We know Flux$$=\dfrac{q}{\epsilon_0}$$
Since $$q=0$$, flux$$=0$$
Flux through the circular part+flux through the hemispherical part$$=0$$
Flux through the hemispherical part$$=-$$Flux through the circular part
Flux through the hemispherical part is
$$\phi=\bar{E}.d\bar{s}$$
$$\phi=|-E(Area \quad of \quad circular \quad region)|$$
$$\phi={E. \pi r^2}$$
Option A is correct
Question 8
1 / -0

There are three concentric thin sphere of radius a, b, c ( a > b > c ) . the total surface charges densities on their surfaces are $$ \sigma , - \sigma , \ sigma $$ respectively. the magnitude of electric field at r ( distance from centre ) such that a > r > b is :

A
0

B
$$ \dfrac { \sigma }{ \varepsilon_0 r^2 } (b^2-c^2) $$

C
$$ \dfrac { \sigma }{ \varepsilon_0 r^2 } (a^2+b^2) $$

D
none of these

Solution

Sphere of radius $$a,b,c$$ $$a>b>c$$ densities of charges surface $$=\sigma ,-\sigma $$ sigma respectively.
The magnitude of electric field at $$=r$$ (distance from centre)
$$a>r>b$$
Using Gauss law,
$$E=\dfrac { \sigma }{ { \epsilon }_{ 0 }{ r }^{ 2 } } \left( { a }^{ 2 }+{ b }^{ 2 } \right) $$
$${ V }_{ A }={ V }_{ C }$$
$$\left( a-b+c \right) =\dfrac { { a }^{ 2 }-{ b }^{ 2 } }{ c } +C$$
$$a-b=\dfrac { { a }^{ 2 }+{ b }^{ 2 } }{ r } $$
Now,
$$E=\dfrac { \sigma }{ { \epsilon }_{ 0 }{ r }^{ 2 } } \left( { a }^{ 2 }+{ b }^{ 2 } \right) $$
Question 9
1 / -0

Two charges of equal magnitudes and at a distance $$r$$ exert a force $$F$$ on each other. If the charges are halved and the distance between them is doubled, then the new force acting on each charge is

A
$$ \dfrac{ F }{ 8 } $$

B
$$ \dfrac{ F }{ 4 } $$

C
$$ \dfrac{ F }{ 16 } $$

D
$$4F$$

Solution

Correct Answer: Option C.

Hint: Force acting between two charges is $$F = \dfrac{{K{q_1}{q_2}}}{{{r^2}}}$$.
Explanation of Correct Option:
Step 1: Required Formula
Force acting between two charges $$F = \dfrac{{K{q_1}{q_2}}}{{{r^2}}}$$.
Charges are of equal magnitude $${q_1} = {q_2} = q$$.
Step 2: Find the force exerted by two charges on each other
Force acting between two charges $$F = \dfrac{{K{q_1}{q_2}}}{{{r^2}}}$$.
Substitute $${q_1} = {q_2} = q$$
As the charges are of equal magnitude the force between the charges is
$$F = \dfrac{{Kqq}}{{{r^2}}}$$
$$F = \dfrac{{K{q^2}}}{{{r^2}}}$$
Step 3: Find the New force exerted by two charges on each other
Magnitude of the charges is $${q_1'} = {q_2'} = \dfrac{q}{2}$$
Distance between the charges $$r' = 2r$$
Force acting between two charges is $${F'} = \dfrac{{Kq_1'q_2'}}{{{r'}^2}}$$
$${F'} = \dfrac{{K\dfrac{q}{2}\dfrac{q}{2}}}{{{{\left( {2r} \right)}^2}}}{\mkern 1mu}$$
$${F'} = \dfrac{{K\dfrac{{{q^2}}}{4}}}{{4{r^2}}}$$
$${F'} = \dfrac{{K{q^2}}}{{16{r^2}}}$$
$${F'} = \dfrac{F}{{16}} $$
Hence the New force exerted by two charges on each other is $$\dfrac{F}{{16}}$$.
Question 10
1 / -0

A charged oil drop is suspended in a uniform field of 3 x 10⁴ v/m so that it neither falls nor rises. The charge on the drop will (Take the mass of the charge = 9.9 x 10^-15 kg and g = 10 m/s² )

A
$$3.3\times 10^{-18} N/C$$

B
$$2.25\times 10^{10} N/C$$

C
$$5\times 10^{10} N/C$$

D
None

Solution

$$\begin{array}{l} At\, \, equilibrium,electric\, \, force\, \, of\, \, drop\, \, balances\, \, weight\, \, of\, drop. \\ qE=mg \\ q=\dfrac { { mg } }{ E } \\ Now, \\ q=\dfrac { { 9.9\times { { 10 }^{ -15 } } } }{ { 3\times { { 10 }^{ 4 } } } } \\ =3.3\times { 10^{ -18 } }C \end{array}$$