Electric Charges and Fields Test

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Electric Charges and Fields Test - 51

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Question 1
1 / -0

The force of repulsion between two point charges is $$F,$$ when these are $$1{\text{ }}m$$ apart. Now the point charges are replaced by conducting spheres of radii $$5{\text{ }}cm$$ having the charge same as that of point charges. The distance between their centres is $$1{\text{ }}m,$$ then the force of repulsion will:

A
increase

B
decrease

C
remain same

D
become $$\dfrac{{10F}}{9}$$

Solution

From Coulombs law
$$F\alpha\dfrac{q_{1}q_{2}}{r^{2}}$$
As the charges and distance are same in either cases, the force between the charges remains constant.
Hence, the force of repulsuion in the second case equal to $$F$$
Question 2
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A metal plate of surface area $$2{m^2}$$ is charged with $$\sqrt {8.85} \,\mu C.$$ What is the mechanical force acting on the plate if it is kept in air? $$\left[ {{\varepsilon _0} = 8.85 \times {{10}^{ - 12}}{C^2}} \right]$$

A
$$0.5 N$$

B
$$0.25 N$$

C
$$1 N$$

D
$$1.5 N$$

Solution

$$E=\dfrac{Q}{A\varepsilon_0}$$
$$E=\sqrt{8.85 \times 10^{-6}}$$
Mechanical force $$=\dfrac{1}{2} \varepsilon_0 E^2 \times A$$
$$=\dfrac{1}{2} \varepsilon_0 \dfrac{Q^2\times A}{A^2 \varepsilon_0^2}$$
$$f=\dfrac{1}{2} \dfrac{Q^2}{A\varepsilon_0}$$
$$f=\dfrac{1}{2} \dfrac{8.85 \times 10^{-6}}{2\times 8.85\times 10^{-12}}=\dfrac{1}{4}$$
$$F \cong 0.25 N$$
Question 3
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Two positive point charges are placed at the distance $$a$$ apart have sum $$Q$$. What values of the charges, coulomb force between them is maximum

A
$$q_1\ = \ q_2\ = \large{\frac{Q}{2}}$$

B
$$q_1\ = \large{\frac{3Q}{4}}$$ , $$ q_2\ = \large{\frac{Q}{4}}$$

C
$$q_1\ = \large{\frac{5Q}{6}}$$ , $$ q_2\ = \large{\frac{Q}{6}}$$

D
None of the above

Solution

net the two charges be $${ q }_{ 1 }$$ & $${ q }_{ 2 }$$
$$\rightarrow { q }_{ 1 }+{ q }_{ 2 }=Q$$
Force between them is
$$F=\dfrac { K{ q }_{ 1 }{ q }_{ 2 } }{ { a }^{ 2 } } $$
The force between them will be maximum only
when $${ q }_{ 1 }={ q }_{ 2 }$$
then, $${ q }_{ 1 }+{ q }_{ 1 }=Q$$
$${ q }_{ 1 }=\dfrac { Q }{ 2 } $$
$$\therefore { q }_{ 1 }={ q }_{ 2 }=\frac { Q }{ 2 } $$
Hence option (A) is correct
Question 4
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A uniform line charge with linear density $$\lambda$$ lies along the y-axis. What flux crosses a spherical surface centred at he origin with $$r\ =\ R$$

A
$$\large{\frac{2R\lambda}{\epsilon_0}}$$

B
$$\large{\frac{R\lambda}{\epsilon_0}}$$

C
$$\large{\frac{\lambda}{\epsilon_0}}$$

D
none of the above

Solution

Line charge density $$=\lambda $$
radius, $$r=R$$
Now, $$\lambda dl=d\theta $$
for total charge enclosed by radius R, $$\int _{ O }^{ R }{ Q } =\int _{ O }^{ R }{ \lambda dl } $$
$$\boxed { { Q }_{ inc }=\lambda R } $$
Now, $$\phi =\dfrac { { Q }_{ inc } }{ { \varepsilon }_{ 0 } } $$ {from Gauss law}
$$=\frac { \lambda R }{ { \varepsilon }_{ 0 } } $$ (Option B is the correct option)
Question 5
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When two spheres having $$2$$Q and $$-$$Q are placed at a certain distance, the force acting between them is F. Then they are connected by a conducting wire and again separated from each other. How much force will act between them if the separation now is the same as before?

A
F

B
$$\dfrac{F}{2}$$

C
$$\dfrac{F}{4}$$

D
$$\dfrac{F}{8}$$

Solution

According to Coulomb's law, force between two charge particle $$q_{1}$$ and $$q_{2}$$ at a distance r is $$\dfrac{Kq_{1}q_{2}}{r^{2}}$$.
Given, Initial Force(magnitude) = $$\dfrac{2KQ^2}{r^{2}}$$ = F.
But, when they are connected by a wire, total charge i.e 2Q +(-Q) = Q will distribute equally (Q/2 each).
Now the force will be $$\dfrac{KQ^2}{2 \times 2\times r^{2}}$$ = $$\dfrac{KQ^2}{4r^{2}}$$ =$$ \dfrac{F}{8} $$.
Therefore, D is the correct option.
Question 6
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Two identical balls carrying charges $$ + 5\mu C$$ and $$ - 2\mu C$$ attracts each other with force $$F$$ in air. After keeping them into contact they are placed at same distance in water $$\left( {{ \in _r} = 81} \right)$$ the new force will be

A
$$\dfrac{9}{{40}}F$$

B
$$F$$

C
$$\dfrac{{F}}{{81}}$$

D
$$\dfrac{F}{{360}}$$

Solution

Two identical balls carrying charges $$=+5\mu e$$ and $$-2\mu e$$
Force $$=F$$ (is air)
$${ \in }_{ r }=81$$
Using coulomb's law,
$$F=\dfrac { 1 }{ 4\pi { \in }_{ 0 } } \dfrac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } $$
$$F=\dfrac { 1 }{ 4\pi { \in }_{ 0 } } \times \dfrac { 5\times 2 }{ { r }^{ 2 } } $$
$${ F }^{ 1 }=\dfrac { 1 }{ K4\pi { \in }_{ 0 } } \times \dfrac { 5\times 2 }{ { r }^{ 2 } } =\dfrac { 1 }{ 81 } \times \left( F \right) $$
or, $${ F }^{ 1 }=\dfrac { F }{ 81 } $$.
Question 7
1 / -0

The magnitude of the average electric field normally present in the atmosphere just above the surface of the Earth is about $$150\ N/C$$, directed inward towards the center of the Earth. This gives the total net surface charge carried by the Earth to be (approximately):
$$\left[Given\ { \varepsilon }_{ 0 }=8.85\times { 10 }^{ -12 }{ C }^{ 2 }/{ N-m }^{ 2 },{ R }_{ E }=6.37\times { 1 }0^{ 6 }m \right]$$

A
$$+\ 670\ kC$$

B
$$-\ 670\ kC$$

C
$$-\ 680\ kC$$

D
$$+\ 680\ kC$$

Solution

Earth $$=150N/C$$
$${ \epsilon }_{ 0 }=8.85\times { 10 }^{ -12 }{ C }^{ 2 }/N-{ m }^{ 2 }$$
$${ R }_{ E }=6.37\times { 10 }^{ 6 }m$$
Given, Electric field $$E=150N/C$$
Total surface charge carried by earth $$q=?$$
According to Gauss's law
$$\phi =\dfrac { q }{ { \epsilon }_{ 0 } } =EA$$
$$q={ \epsilon }_{ 0 }EA$$
$$={ \epsilon }_{ 0 }E\pi { r }^{ 2 }$$
$$=8.854\times { 10 }^{ -12 }\times 150\times { \left( 6.37\times { 10 }^{ 6 } \right) }^{ 2 }$$
$$=680KC$$
Question 8
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If the intensity of electric field at a distance $$x$$ from the centre in axial position of small electric dipole is equal to the intensity at a distance $$y$$ in equatorial position, then

A
$$x = y$$

B
$$x = y/2$$

C
$$y = x/2^{2/3}$$

D
$$y = x/2^{1/3}$$

Solution

Electric field due to dipole at axial point
$$E_x=\dfrac{2KP}{x^3}$$
Electric field at equilibrial point $$E_y=\dfrac{KP}{y^3}$$
$$E_x=E_y\Rightarrow \dfrac{2}{x^3}=\dfrac{1}{y^3}$$
$$\Rightarrow y=\dfrac{x}{2^{1/3}}$$
Question 9
1 / -0

Two particles each of mass $$m$$ and charge $$q$$ are separated by $$r_1$$ and the system is left free to move at $$t = 0$$. At $$t = t$$, both the particles are found to be separated by $$r_2$$. The speed of each particles is

A
$$\dfrac{qm}{4\pi \varepsilon_0 r_1r_2}$$

B
$$\dfrac{q}{r_1r_2\sqrt{(r^2_2 - r^2_1)/4\pi \, \varepsilon_0m}}$$

C
$$\dfrac{\sqrt{2}q}{r_1r_2\sqrt{(r^2_2 - r^2_1)/4\pi \, \varepsilon_0m}}$$

D
$$q\sqrt{\dfrac{(r_2 - r_1)}{4\pi \varepsilon_0 mr_1r_2}}$$

Solution

Since the particles are of same mass thus each particle will have same speed.
Initial energy, $$E_1 = \dfrac{1}{4\pi \varepsilon_0}\dfrac{q^2}{r_1} + 0$$

Final energy, $$E_t = \dfrac{1}{4\pi \varepsilon_0}\dfrac{q^2}{r^2} + \dfrac{1}{2}mv^2 + \dfrac{1}{2}mv^2 = \dfrac{1}{4\pi \varepsilon_0}\dfrac{q^2}{r^2} + mv^2$$

As the field is conservative,
Hence applying COE,
$$\dfrac{1}{4\pi \varepsilon_0}\dfrac{q^2}{r_1} = \dfrac{1}{4\pi \varepsilon_0}\dfrac{q^2}{r_2} + mv^2$$

$$\therefore mv^2 = \dfrac{1}{4\pi \varepsilon_0}\left(\dfrac{1}{r_1} - \dfrac{1}{r_2}\right)$$

$$\therefore v = q \sqrt{\dfrac{(r_2 - r_1)}{4\pi \varepsilon_0 mr_1r_2}}$$.
Question 10
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A particle of mass $$100 gm$$ and charge $$2 \mu C$$ is released from a distance of $$50 cm$$ from a fixed change of $$5\mu C$$. Find the speed of the particle when its distance from the fixed charge becomes $$3 m$$.

A
$$\sqrt{6}m/s$$.

B
$$\sqrt{3}m/s$$.

C
$$\sqrt{4}m/s$$.

D
$$\sqrt{5}m/s$$.

Solution

Loss of potential energy $$=$$ gain in kinetic energy
$$U_1 - U_2 = \Delta K$$.
$$kQq\left[\dfrac{1}{r_1} - \dfrac{1}{r_2}\right] = 1/2 mv^2 - 0$$

$$v = \sqrt{\dfrac{2lQq}{m}\left[\dfrac{r_2}{r_1} - \dfrac{r_1}{r_2}\right]} = \sqrt{\dfrac{2 \times 9 \times 10^9 \times 5 \times 10^{-6} \times 2 \times 10^{-6} \times 2.5}{0.1 \times 3 \times 0.5}} = \sqrt{3}m/s$$.

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Re-Attempt Weekly Quiz Competition

Electric Charges and Fields Test - 51

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Electric Charges and Fields Test - 51

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SHARING IS CARING

Question 1

increase

decrease

remain same

become $$\dfrac{{10F}}{9}$$

Solution

Question 2

$$0.5 N$$

$$0.25 N$$

$$1 N$$

$$1.5 N$$

Solution

Question 3

$$q_1\ = \ q_2\ = \large{\frac{Q}{2}}$$

$$q_1\ = \large{\frac{3Q}{4}}$$ , $$ q_2\ = \large{\frac{Q}{4}}$$

$$q_1\ = \large{\frac{5Q}{6}}$$ , $$ q_2\ = \large{\frac{Q}{6}}$$

None of the above

Solution

Question 4

$$\large{\frac{2R\lambda}{\epsilon_0}}$$

$$\large{\frac{R\lambda}{\epsilon_0}}$$

$$\large{\frac{\lambda}{\epsilon_0}}$$

none of the above

Solution

Question 5

F

$$\dfrac{F}{2}$$

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

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

Solution

Question 6

$$\dfrac{9}{{40}}F$$

$$F$$

$$\dfrac{{F}}{{81}}$$

$$\dfrac{F}{{360}}$$

Solution

Question 7

$$+\ 670\ kC$$

$$-\ 670\ kC$$

$$-\ 680\ kC$$

$$+\ 680\ kC$$

Solution

Question 8

$$x = y$$

$$x = y/2$$

$$y = x/2^{2/3}$$

$$y = x/2^{1/3}$$

Solution

Question 9

$$\dfrac{qm}{4\pi \varepsilon_0 r_1r_2}$$

$$\dfrac{q}{r_1r_2\sqrt{(r^2_2 - r^2_1)/4\pi \, \varepsilon_0m}}$$

$$\dfrac{\sqrt{2}q}{r_1r_2\sqrt{(r^2_2 - r^2_1)/4\pi \, \varepsilon_0m}}$$

$$q\sqrt{\dfrac{(r_2 - r_1)}{4\pi \varepsilon_0 mr_1r_2}}$$

Solution

Question 10

$$\sqrt{6}m/s$$.

$$\sqrt{3}m/s$$.

$$\sqrt{4}m/s$$.

$$\sqrt{5}m/s$$.

Solution

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