Electric Charges and Fields Test

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Electric Charges and Fields Test - 52

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Question 1
1 / -0

Consider an electric field $$\bar { E } =E_{ 0 }\hat { x }$$ where $$E_{0}$$ is a constant. The flux through the shaded area (as shown in the figure) due to this field is

A
$$2E_{0}a^{2}$$

B
$$\sqrt{2}E_{0}a^{2}$$

C
$$E_{0}\ a^{2}$$

D
$$\dfrac{E_{0}a^{2}}{\sqrt{2}}$$

Solution

on locating the plate is tilled lly $$45^o$$
$$\phi=E.S$$
$$=E_0a^2\cos 45$$
$$=\dfrac{E_0a^2}{\sqrt 2}$$
Question 2
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If two like charges of magnitude $$1\times {10}^{-9}$$ coulomb and $$9\times {10}^{-9}$$ coulomb are separated by a distance of $$1$$ meter, then the point on the line joining the charges, where the force experinced by a charge placed at that point is zero, is:

A
$$0.25\ m$$ from the charge $$1\times {10}^{-9}$$ coulomb

B
$$0.75\ m$$ from the charge $$1\times {10}^{-9}$$ coulomb

C
Both $$1$$ and $$2$$

D
at all points on the line joining the charges

Solution

Two like charges are of magnitude $$Q_1=1 \times 10^{-9}$$ coulomb and $$Q_2=9 \times 10^{-9}$$ coulomb respectively.
Let the point at which the force is zero be at $$x$$ metre distance from charge $$Q_1$$
So, distance from $$Q_2$$ is $$(1-x)$$ metre.
Force due to $$Q_1$$ = Force due to $$Q_2$$

$$\begin{array}{l} \dfrac { { K\times 1\times { { 10 }^{ -9 } } } }{ { { x^{ 2 } } } } =\dfrac { { K\times 9\times { { 10 }^{ -9 } } } }{ { { { \left( { 1-x } \right) }^{ 2 } } } } \\ { \left( { 1-x } \right) ^{ 2 } }=9{ x^{ 2 } } \\ x=\dfrac { 1 }{ 4 } m. \end{array}$$

i.e. $$0.25$$m from the charge $$1 \times 10^{-9}$$C.
Question 3
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Three equal charges (q) are placed at corners of equilateral triangle . The force on any charge is

A
zero

B
$$\sqrt{3}\dfrac{Kq^2}{a^2}$$

C
$$\dfrac{Kq^2}{\sqrt{3}a^2}$$

D
3$$\sqrt{3}\dfrac{Kq^2}{a^2}$$

Solution

Let, find out force on charge at $$A$$.
$$\therefore \quad { \vec { F } }_{ AB }=\dfrac { K\left( q \right) \left( q \right) }{ { a }^{ 2 } } =\dfrac { { Kq }^{ 2 } }{ { a }^{ 2 } } $$
Similarly,
$${ \vec { F } }_{ AC }=\dfrac { K{ q }^{ 2 } }{ { a }^{ 2 } } $$
Now, Angle b/w $${ \vec { F } }_{ AB }\quad \& \quad { \vec { F } }_{ AC }={ 60 }^{ 0 }$$
Using parallelogram of vector addition,
$${ \vec { F } { r }_{ 1 } }=\sqrt { { \left( { \vec { F } }_{ AB } \right) }^{ 2 }+{ \left( { \vec { F } }_{ AC } \right) }^{ 2 }+2{ \vec { F } }_{ AB }{ \vec { F } }_{ AC }\cos\theta } $$
$$=\sqrt { \left( 3 \right) { \left( \dfrac { { Kq }^{ 2 } }{ { a }^{ 2 } } \right) }^{ 2 } } $$
$$=\dfrac { K{ q }^{ 2 } }{ { a }^{ 2 } } \sqrt { 3 } N$$
All the three charges will experience the same force but in different dir.

$$\therefore $$ Option (B) is the correct answer.
Question 4
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Two small identical spheres having charges $$+10\mu C$$ and $$- 90\mu C$$ attract each other with a force of $$F$$ Newton. If they are Kept in contact and then separated by the same distance, The new force between them is:

A
$$F/6$$

B
$$16/F$$

C
$$16F/9$$

D
$$9F$$

Solution

Initial force,
$$F = k\cfrac{900 \times 10^{-12}}{r^2} N$$
After keeping in contact, due to induction;

$$q_A = -40\mu C$$

$$q_B = -40\mu C$$

So, new force;
$$F' = k\cfrac{1600 \times 10^{-12}}{r^2} N$$

$$\cfrac{F'}{F} = \cfrac{1600}{900}$$

$$F' = \cfrac{16}{9} F$$
Question 5
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A charge particle $$q$$ is shot from a large distance with speed $$v$$ towards a fixed charge particle $$Q$$. It approaches $$Q$$ upto a closest distance $$r$$ and then turns. If $$q$$ were given a speed $$2v$$ the closest distance of approach would be:

A
$$r$$

B
$$2r$$

C
$$\dfrac {r}{2}$$

D
$$\dfrac {r}{4}$$

Solution

According to question,
When speed is $$v$$.
$$\frac{1}{2}mv^{2}=\frac{KQq}{r}-------------------(1)$$
When speed is $$2v$$.
$$\frac{1}{2}m{4v}^2=\frac{KQq}{r'}--------------(2)$$
Divide equation $$1$$ with equation $$2$$-
$$4r'=r$$
$$r'=\frac{r}{4}$$
Question 6
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Two equal and like charges when placed 5 cm apart experience a repulsive force of 0.144 newton. The magnitude of the charge in microcoloumb will be :

A
0.2

B
2

C
20

D
12

Solution

The force is given from Coulomb’s law as,

$$F = \frac{1}{{4\pi {\varepsilon _0}}} \times \frac{{{q_1}{q_2}}}{{{r^2}}}$$

$$0.144 = \frac{{9 \times {{10}^9} \times q \times q}}{{{{\left( {0.05} \right)}^2}}}$$

$$q = 0.2 \times {10^{ - 6}}\;{\rm{C}}$$

$$ = 0.2\;{\rm{\mu C}}$$

Thus, the magnitude of the charge in micro coulomb is $$0.2$$.
Question 7
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Three charge +4q, Q and q are placed in a straight line of length $$l$$ at points distance 0, $$\dfrac{l}{2}$$ and $$l$$ respectively. What should be the value of Q in order to make the net force on q to be zero?

A
$$-q$$

B
$$-2q$$

C
$$-q/2$$

D
$$4q$$

Solution

Two force are acting on $$q$$ one due to $$4q$$ and second due to $$Q$$
$$F_1=\cfrac{1}{4\pi\epsilon_0}\cfrac{4q^2}{12}\\F_2=\cfrac{1}{4\pi\epsilon_0}\cfrac{Qq}{(1/2)^2}\\ \quad=\cfrac{1}{4\pi\epsilon_0}\cfrac{(4Q)q}{12}\\F_1+F_2=0\\ \cfrac{1}{4\pi\epsilon_0}\cfrac{(4q)q}{12}+\cfrac{1}{4\pi\epsilon_0}\cfrac{(4Q)q}{12}=0\\Q=-q$$
Question 8
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A point charge Q is placed at origin O. Let $$\overrightarrow {{E_A}} $$,$$\overrightarrow {{E_B}} $$ and $$\overrightarrow {{E_C}} $$ represent electric fields at A, B and C respectively. If coordination of A,B and C are respectively (1,2,3) m,(1,1,-1) m and (2,2,2) m then

A
$$\overrightarrow {{E_A}} \bot \overrightarrow {{E_B}} $$

B
$$\overrightarrow {{E_A}} \parallel \overrightarrow {{E_B}} $$

C
$$\left| {\overrightarrow {{E_B}} } \right|\parallel 4\left| {\overrightarrow {{E_C}} } \right|$$

D
$$\left| {\overrightarrow {{E_B}} } \right|\parallel 8\left| {\overrightarrow {{E_C}} } \right|$$

Solution

$$\overrightarrow{E_a}=\cfrac{h_E}{ra^3}=\cfrac{h_E}{1^2+2^2+3^2}(1\hat{i}+2\hat{j}+3\hat{k})\\ \overrightarrow{E_a}=\cfrac{h_E}{14^{3/2}}(1\hat{i}+2\hat{j}+3\hat{k})\\ \overrightarrow{E_b}=\cfrac{ha}{r_b^2}\overrightarrow{r_b}=\cfrac{h_E}{(1^2+1^2+1^2)}^{3/2}(1\hat{i}-1\hat{j}+1\hat{k})\\=\cfrac{h_E}{3^{3/2}}(1\hat{i}-1\hat{j}+1\hat{k})\\ \overrightarrow{E_c}=\cfrac{h_E}{r_c^2}\overrightarrow{r_c}=\cfrac{h_E}{(2^2+2^2+2^2)^{3/2}}(2\hat{i}+2\hat{j}+2\hat{k})$$
$$\quad=\cfrac{h_E}{12^{3/2}}(2\hat{i}+2\hat{j}+2\hat{k})$$
Now
$$\overrightarrow{E_a}.\overrightarrow{E_b}=\cfrac{h_E}{14^{3/2}}(1\hat{i}+2\hat{j}+3\hat{k})=\cfrac{h_E}{3^{3/2}}(1\hat{i}-1\hat{j}+1\hat{k})\\ \Rightarrow \overrightarrow{E_a}.\overrightarrow{E_b}=(\cfrac{h_E}{14^{3/2}})(\cfrac{h_E}{3^{3/2}})(1-2+3)\neq0$$
Thus$$\overrightarrow{E_a}$$ and $$\overrightarrow{E_b}$$ are perpendicular to each other
$$|E_c|=\cfrac{h_E}{12^{3/2}}(2^2+2^2+2^2)^{1/2}=\cfrac{h_E}{12^{3/2}}(12)^{1/2}\\ \Rightarrow |\overrightarrow{E_c}|=\cfrac{h_E}{12}\\ \overrightarrow|E_b|=\cfrac{ha}{3^{3/2}}(1^2+1^2+1^2)^{1/2}=\cfrac{_E}{3}=4\times\cfrac{h_E}{12}=4|\overrightarrow{E_c}|$$
So, $$|\overrightarrow{E_b}|=4|E_c|$$
Question 9
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A point charge is placed at the corner of a cube. The electric flux through the shaded surface is

A
$${q \over {8{\varepsilon _0}}}$$

B
$${q \over {{\varepsilon _0}}}$$

C
$${q \over {24{\varepsilon _0}}}$$

D
$${q \over {12{\varepsilon _0}}}$$

Solution

$$\textbf{Step 1: Choosing gaussian surface}$$
To calculate the electric flux through the shaded area, we have to make a Gaussian surface such that a given charge appears at the center of the bigger cube resulting from the combination of $$8$$ small cubes (as shown in figure)

$$\textbf{Step 2: Flux through one cube}$$
If a charge is at the corner then it is being shared by $$8$$ cubes.
Flux passing through the given cube will be $$= \dfrac{q}{8\epsilon_0}$$

$$\textbf{Step 3: Electric flux through shaded area}$$
Now for any cube, it is touching $$3$$ of its faces. The area vector of that side and electric field vector will be perpendicular so flux through $$3$$ sides will be $$0$$.
An equal amount of flux will flow from other $$3$$ sides.
Flux through shaded area $$= \dfrac{q}{8 \times 3 \in_0}$$
$$= \dfrac{q}{24 \in_0}$$

Hence, Option(C) is correct.
Question 10
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Charge Q is uniformly distributed over a ring of radius r .Electric field at a point on the axis is maximum at a distance x from the centre. x is equal to

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

B
$$\dfrac{r} {2}$$

C
$$\sqrt 2 r$$

D
$$r$$

Solution

The electric field due to the ring on its axis at a distance x is given by:-
$$ E = \dfrac{kqx}{(x^2 + R^2)^{3/2}}$$
To find maximum electric field, we will use the concept of maximum and minimum :-
$$ \dfrac{dE}{dx} = kq \dfrac{(x^2 + R^2)^{3/2}-3/2 (x^2 + R^2 )^{1/2}.2x^2}{(x^2 + R^2)}$$
Now, $$ \dfrac{dE}{dx} = 0 \Rightarrow (x^2 + R^2)^{3/2} = \dfrac{3}{2} \times 2x^2 (x^2 + R^2)^{1/2}$$
$$ \Rightarrow x^2 + R^2 = 3x^2 $$
$$ \Rightarrow 2x^2 = R^2 $$
$$ \Rightarrow x^2 = \dfrac{R^2}{2}$$
$$ \Rightarrow x = \pm \dfrac{R}{\sqrt{2}}$$