Electric Charges and Fields Test

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Electric Charges and Fields Test - 64

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Weekly Quiz Competition

Question 1
1 / -0

An electron with a speed of $$4×10^6$$ m/s enters an electric field of magnitude $$10^5$$N/C traveling along field lines in the direction that retards its motion. Calculate the minimum length of the electric field region in the direction of motion required to stop the electron momentarily within the field region.

A
$$4.49m$$

B
$$3.42m$$

C
$$4.15m$$

D
$$1.245m$$

Solution

Initial velocity, u = $$4×10^6$$m/s

Electric field strength, E = $$10^5$$ N/C

Retardation of the electron can be given as:
$$a=\dfrac{F}m= \dfrac{qE}m =1.76 ×10^{12}$$ units
Where,
q = charge of an electron= $$1.6×10^{-19}$$ C
m = mass of electron= $$9.1×10^{-31}$$

From equation of motion we have:
$$v^2$$= $$u^2$$+$$ 2as $$

So, Putting all the respective values we get:
$$0$$ = $$16×10^{12}$$ - $$3.56×10^{12}s$$

$$s=4.49$$m

Where $$s$$ is the required length of the field region.
Question 2
1 / -0

A ball of charge to mass ratio $$8 \mu C/g$$ is placed at a distance of $$10 cm$$ from the wall. An electric field $$100\ N/m$$ is switched on in the direction of wall. Find the time period of its oscillations. Assume all collisions elastic.

A
$$1\sec$$

B
$$2\sec$$

C
$$3\sec$$

D
$$4\sec$$

Solution

$$a = \dfrac {qE}{m} = \dfrac {8\times 10^{-6}}{10^{-3}} \times 100 = 0.8\ m/s^{2}$$

As electric field is switched $$ON$$, ball first strikes to wall and returns back.
One oscillation
Thus
$$s = ut + \dfrac {1}{2} at_{1}^{2}$$
$$0.1 = \dfrac {1}{2} \times 0.8t_{1}^{2}$$
$$t_{1} = \dfrac {1}{2} \sec$$

Thus, time period
$$T = 2\times \dfrac {1}{2} = 1\sec$$
Question 3
1 / -0

A charge $$Q$$ is divided into $$q$$ and $$(Q - q)$$. If $$\dfrac {Q}{q} = x$$, such that the repulsion between them is maximum, find $$x$$.

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

As we know, $$F = \dfrac {k(Q - q)q}{d^{2}}$$
For $$F$$ to be maximum,
$$\dfrac {dF}{dq} = 0$$
$$\Rightarrow Q - 2q = 0$$
$$\Rightarrow \dfrac {Q}{q} = 2$$
$$\Rightarrow x = 2$$
Question 4
1 / -0

A hemispherical body is placed in a uniform electric field E. What is the flux linked with curved surface, if field is perpendicular to base in figure.

A
$$2\pi R^2E$$

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

C
$$\pi R^2E$$

D
Zero

Solution

The flux entering the hemisphere is equal to the flux leaving it,
Thus,
$$ \phi_{in} = \phi_{out} $$
$$ \phi_{out} = \pi R^{2} E $$
Question 5
1 / -0

Consider a uniform electric field $$E=3\times 10^3 \hat i \: N/C$$. What is the flux of this field through a square of $$10\ cm$$ on a side whose plane is parallel to the yz plane?

A
$$30 \:N m^2/C$$

B
$$40 \:N m^2/C$$

C
$$50 \:N m^2/C$$

D
$$60 \:N m^2/C$$

Solution

The flux of an electric field is given by,
$$ \phi = EA$$
$$ \phi = 3 \times {10}^{3} \times 0.1 \times 0.1 $$
$$ \phi = 30\ Nm^2/C $$
Question 6
1 / -0

In the figure shown, A is a fixed charged. B (of mass m) is given a velocity $$v$$ as shown in figure. At this moment the radius of curvature of the resultant path of B is :

A
$$\dfrac{8 \pi \varepsilon_0 r^2 mv^2}{\sqrt{3}q^2}$$

B
$$\dfrac{3 \pi \varepsilon_0 r^2 mv^2 }{q^2}$$

C
$$\dfrac{4 \pi \epsilon_0 r^2 mv^2}{q^2}$$

D
$$r$$

Solution

$$F=\dfrac{kq^2}{r^2}=\dfrac{mv^2}{R_c}$$

$$\Rightarrow R_c=\dfrac{mv^2}{F}=\dfrac{mv^2}{\dfrac{kq^2}{r^2}}=\dfrac{4\pi \varepsilon_0 r^2 mv^2}{q^2}$$
Question 7
1 / -0

A particle of charge $$-q$$ and mass $$m$$ moves in a circular orbit of radius $$r$$ about a fixed charge $$+Q$$. The relation between the radius of the orbit $$r$$ and the time period $$T$$ is:

A
$$r=\dfrac{Qq}{16\pi ^{2}\epsilon _{0}m}T^{3}$$

B
$$r^{3}=\dfrac{Qq}{16\pi ^{3}\epsilon _{0}m}T^{2}$$

C
$$r^{2}=\dfrac{Qq}{16\pi ^{3}\epsilon _{0}m}T^{3}$$

D
$$r^{2}=\dfrac{Qq}{16\pi\epsilon _{0}m}T^{3}$$

Solution

The centrifugal force acting on charge $$q$$ is:
$$F=\dfrac {k q Q}{r^2}$$
$$\Rightarrow \dfrac {mv^2}{r}=\dfrac {k q Q}{r^2}$$
$$\Rightarrow V=\sqrt {\dfrac {k q Q}{rm}}$$
Now, $$V=\dfrac {2\pi r}{T}$$
$$\Rightarrow \dfrac {2\pi r}{T}=\sqrt {\dfrac {q Q}{4\pi \varepsilon_0 r m}}$$
$$\Rightarrow \dfrac {4\pi^2 r^2}{T^2}=\dfrac {q Q}{4\pi \varepsilon_0 r m}$$
$$\Rightarrow r^3=\dfrac {q Q T^2}{16\pi^3 \varepsilon_0 m}$$
Question 8
1 / -0

Two point charges placed at a distance $$r$$ in air exert a force $$F$$ on each other. The value of distance $$R$$ at which they experience force $$4F$$ when placed in a medium of dielectric constant $$K = 9$$ is :

A
$$r$$

B
$$\dfrac{r}{6}$$

C
$$\dfrac{r}{8}$$

D
$$2r$$

Solution

Initially force, $$F=\dfrac{1}{4\pi\epsilon_0}\dfrac{q_1q_2}{r^2} .......(1)$$

Finally, $$4F=\dfrac{1}{4\pi\epsilon}\dfrac{q_1q_2}{R^2}$$ where $$K=\dfrac{\epsilon}{\epsilon_0}\Rightarrow \epsilon=9\epsilon_0$$

now,$$4F=\dfrac{1}{4\pi(9\epsilon_0)}\dfrac{q_1q_2}{R^2}......(2)$$
$$(1)/(2) ,\dfrac{1}{4}=\dfrac{9R^2}{r^2}\Rightarrow R=\dfrac{r}{6}$$
Question 9
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A positive point charge $$+Q$$ is fixed in space. A negative point charge $$q$$ of mass $$m$$ revolves around the fixed charge in an elliptical orbit. The fixed charge $$+Q$$ is at one focus of the ellipse. The only force acting on the negative charge is the electrostatic force due to the positive charge. Then which of the following statements is true?

A
Linear momentum of negative point charge is conserved.

B
Angular momentum of negative point charge about the fixed positive charge is not conserved.

C
Angular momentum of negative point charge about the fixed positive charge is conserved.

D
Angular velocity of the charge $$q$$ is conserved about $$+Q$$ charge.

Solution

Since net force on the negative charge is always directed towards fixed positive charge, the torque on negative charge about positive charge is zero. Therefore angular momentum of negative charge about the fixed positive charge is conserved.
$$L = (mr^2) \omega $$
$$\therefore$$ As value of r changes, $$ \omega $$ also changes
Question 10
1 / -0

Electric flux through a surface of area $$100\ m^{2}$$ lying in the xy plane is (in V-m) if $$\vec{E}=\hat{i}+\sqrt{2}\hat{j}+\sqrt{3\hat{k}}$$

A
$$100$$

B
$$141.4$$

C
$$173.2$$

D
$$200$$

Solution

Given $$\overrightarrow E= \hat i+ \sqrt 2 \hat j+ \sqrt 3 \hat k$$, Area, $$\overrightarrow A= 100 \hat k$$
Electric Flux$$= \overrightarrow E.\overrightarrow A$$
$$\Rightarrow \text{Electric Flux}= (\hat i + \sqrt 2 \hat j + \sqrt 3 \hat k).( 100 \hat k)=100\sqrt 3=173.2$$

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Re-Attempt Weekly Quiz Competition

Electric Charges and Fields Test - 64

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Electric Charges and Fields Test - 64

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SHARING IS CARING

Question 1

$$4.49m$$

$$3.42m$$

$$4.15m$$

$$1.245m$$

Solution

Question 2

$$1\sec$$

$$2\sec$$

$$3\sec$$

$$4\sec$$

Solution

Question 3

$$1$$

$$2$$

$$3$$

$$4$$

Solution

Question 4

$$2\pi R^2E$$

$$\pi R^2E/2$$

$$\pi R^2E$$

Zero

Solution

Question 5

$$30 \:N m^2/C$$

$$40 \:N m^2/C$$

$$50 \:N m^2/C$$

$$60 \:N m^2/C$$

Solution

Question 6

$$\dfrac{8 \pi \varepsilon_0 r^2 mv^2}{\sqrt{3}q^2}$$

$$\dfrac{3 \pi \varepsilon_0 r^2 mv^2 }{q^2}$$

$$\dfrac{4 \pi \epsilon_0 r^2 mv^2}{q^2}$$

$$r$$

Solution

Question 7

$$r=\dfrac{Qq}{16\pi ^{2}\epsilon _{0}m}T^{3}$$

$$r^{3}=\dfrac{Qq}{16\pi ^{3}\epsilon _{0}m}T^{2}$$

$$r^{2}=\dfrac{Qq}{16\pi ^{3}\epsilon _{0}m}T^{3}$$

$$r^{2}=\dfrac{Qq}{16\pi\epsilon _{0}m}T^{3}$$

Solution

Question 8

$$r$$

$$\dfrac{r}{6}$$

$$\dfrac{r}{8}$$

$$2r$$

Solution

Question 9

Linear momentum of negative point charge is conserved.

Angular momentum of negative point charge about the fixed positive charge is not conserved.

Angular momentum of negative point charge about the fixed positive charge is conserved.

Angular velocity of the charge $$q$$ is conserved about $$+Q$$ charge.

Solution

Question 10

$$100$$

$$141.4$$

$$173.2$$

$$200$$

Solution

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