Electric Charges and Fields Test

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Electric Charges and Fields Test - 33

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Weekly Quiz Competition

Question 1
1 / -0

Two charges are placed in vacuum at a distance 'd' apart. The force between them is 'F'. If a medium of a dielectric constant 4 is introduced between them, the force will be:

A
4F

B
2F

C
F/2

D
F/4

Solution

$$F=\dfrac {2}{4\pi \epsilon_0\epsilon_r}\dfrac {q_1q_2}{Fd^2}$$
$$F\alpha \dfrac {1}{\epsilon_r}$$
For vacuum $$\epsilon_r=1,$$
$$\epsilon_r$$ of medium $$=4$$
$$\therefore$$ F for medium will be F/4.
Question 2
1 / -0

In Coulomb's law, the constant of proportionality K has the units

A
$$N$$

B
$$Nm^2$$

C
$$NC^2/m^2$$

D
$$Nm^2/C^2$$

Solution

force between two charged particles is given by

$$[F= \dfrac{(K\times q_{1}\times q_{2})}{r^{2}}]$$

$$[K=\dfrac{(F\times r^{2})}{(q_{1} \times q_{2})}]$$

Therefore unit of K will be $$[(N \times m^{2})/(coulomb)^{2}]$$
Question 3
1 / -0

If two charges of 1 coulomb are placed 1 km apart, the force between them will be:

A
$$9\times 10^3N$$

B
$$9\times 10^{-3}$$

C
$$1.1\times 10^{-4}N$$

D
$$10^{-6}N$$

Solution

We know that
$$F=\dfrac {9\times 10^9\times q\times q}{(r)^2}$$
$$F=\dfrac {9\times 10^9\times 1\times 1}{(1000)^2}$$
$$9\times 10^{9-6}=9\times 10^3N$$
Question 4
1 / -0

Two positively charged particles each of mass $$1.7 \times 10^{-27}$$ kg and carrying a charge of $$1.6 \times 10^{-19}$$ C are placed at a distance $$r$$ apart. If each one experiences a repulsive force equal to its weight, then $$r$$ is :

A
$$1.17 m$$

B
$$1.71 m$$

C
$$0.117 m$$

D
$$0.171m$$

Solution

The repulsive electrostatic force between the charges is given by,
$$[F = (k\times q^{2})/r^{2}]$$
$$[ F= [9\times 10^{9}\times(1.6\times 10^{-19})^{2}]/r^{2}]$$
it is given that this force is equal to weight of the charge i.e. $$[F =m\times g]$$
$$[9\times10^{9}\times(1.6\times10^{-19})^{2}/r^{2} = 1.7\times10^{-27}\times9.8]$$
$$[r = 0.117]$$ m
Question 5
1 / -0

Cathode rays gain K.E. when they are accelerated by an electric field. If cathode rays are subjected to the action of magnetic field, then their

A
Energy increases

B
Momentum increases

C
Potential energy increases

D
Energy and momentum remain unaffected

Solution

Since the magnetic force acting on a moving electric charge is perpendicular to the motion of the charge - the WORK done on the moving charge is zero, meaning that its kinetic energy stays constant.Since the DIRECTION of the charge’s motion changes, we know that the momentum changes its direction also (but not its magnitude.)
Question 6
1 / -0

A point charge $$+10\mu C$$ is at a distance $$5\ cm$$ directly above the centre of a square of side $$10\ cm$$, as shown in the figure. What is the magnitude of the electric flux through the square?

A
$$1.2\times 10^5 Nm^2/C$$

B
$$1.9\times 10^5 Nm^2/C$$

C
$$3.2\times 10^5 Nm^2/C$$

D
$$4.2\times 10^5 Nm^2/C$$

Solution

Imagine a square cube around the charge with the charge at the centre of the cube and the square being one surface of the cube..
Now apply Gauss law for finding flux through the cube. Divide by 6 to get flux through one face, this is the answer.
$$6\phi =\frac {\phi}{\epsilon_0}$$ where $$\phi =$$ flux through one of the faces of square cube.
$$\phi = 1.9\times 10^5 Nm^2/C$$
Question 7
1 / -0

Find the minimum force between the two electrons around He nucleus. Assume radius of He nucleus $$=$$ 6.8 A$$^o$$.

A
$$12 \times 10^{-10} N$$

B
$$1.2 \times 10^{-10} N$$

C
$$0.12 \times 10^{-10} N$$

D
0.012 N

Solution

The force between the two charges, $$q_1$$ and $$q_2$$ separated by distance $$r$$ is given by,
$$ F = K \dfrac {q_1 q_2}{r^2}$$,
where $$ K = \dfrac {1}{4\pi \varepsilon_{o}} = 9 \times 10^9$$
Since we are considering two electrons of He nucleus, $$q_1 = q_2 = e = 1.6 \times 10^{-19} C$$.
Also the He atom consists of only two electrons. So the distance between them is given by the diameter of the nucleus, i.e., $$r = 2 \times 6.8 \times 10^{10} = 13.6 \times 10^{-10} m $$
Hence,
$$ F = 9 \times 10^9 \dfrac {(1.6 \times 10^{-19}) \times (1.6 \times 10^{-19})}{(13.6 \times 10^{-10})^2}$$ $$ = 1.246 \times 10^{-10} N$$
Question 8
1 / -0

A charged particle moves with a speed v in a circular path of radius r around a long uniformly charged conductor then :

A
$$v \propto r$$

B
$$\displaystyle v \propto \dfrac{1}{r}$$

C
$$\displaystyle v \propto \dfrac{1}{\sqrt{2}}r$$

D
$$v$$ is independent of r

Solution

Consider a Gaussian cynlinder of radius $$r$$ and length $$l$$. Using Gauss's law the electric field at distance $$r$$ is $$\displaystyle E.(2\pi rl) =\dfrac{\lambda l}{\epsilon_0} \Rightarrow E=\dfrac{\lambda}{2\pi\epsilon_0 r}$$

Here the electrostatic force is equal to centrifugal force. i.e,

$$\displaystyle qE=\dfrac{mv^2}{r}$$

$$\displaystyle q\dfrac{\lambda}{2\pi \epsilon_0 r}=\dfrac{mv^2}{r}$$

or$$\displaystyle v^2=\dfrac{q\lambda}{2\pi\epsilon_0 m}$$

so $$v$$ is independent of $$r$$.
Question 9
1 / -0

Two particles having charges $${ q }_{ 1 }$$ and $${ q }_{ 2 }$$ exert a force $$F$$ on each other, when they are placed at a certain distance. What will be the force between them, if the distance between them is reduced to half and the charge on each particle is doubled ?

A
$$12\ F$$

B
$$4\ F$$

C
$$8\ F$$

D
$$16\ F$$

Solution

$$F=k\dfrac{q_1q_2}{r^2}$$

$$F'=k\dfrac{q'_1q'_2}{r'^2}$$

given, $$q'_1=2q_1, q'_2=2q_2, r'=\dfrac{r}{2}$$

$$\therefore \dfrac{F'}{F} =\dfrac{q'_1q'_2}{q_1q_2}\dfrac{r^2}{r'^2}=4\times 4=16$$

$$F'=16 F$$

ans: (D)
Question 10
1 / -0

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

A
$$100$$

B
$$141.4$$

C
$$173.2$$

D
$$200$$

Solution

Electric flux, $$\phi=\int\vec{E}.\hat{n}ds$$

given, $$\hat{n}=\hat{k},ds=100 m^2 \vec{E}=\hat { i } +\sqrt { 2 } \hat { j } +\sqrt { 3 } \hat { k } $$

$$\phi=(\hat { i } +\sqrt { 2 } \hat { j } +\sqrt { 3 } \hat { k }).\hat{k}100=100\sqrt{3}=173.2$$

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Electric Charges and Fields Test - 33

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Electric Charges and Fields Test - 33

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Question 1

4F

2F

F/2

F/4

Solution

Question 2

$$N$$

$$Nm^2$$

$$NC^2/m^2$$

$$Nm^2/C^2$$

Solution

Question 3

$$9\times 10^3N$$

$$9\times 10^{-3}$$

$$1.1\times 10^{-4}N$$

$$10^{-6}N$$

Solution

Question 4

$$1.17 m$$

$$1.71 m$$

$$0.117 m$$

$$0.171m$$

Solution

Question 5

Energy increases

Momentum increases

Potential energy increases

Energy and momentum remain unaffected

Solution

Question 6

$$1.2\times 10^5 Nm^2/C$$

$$1.9\times 10^5 Nm^2/C$$

$$3.2\times 10^5 Nm^2/C$$

$$4.2\times 10^5 Nm^2/C$$

Solution

Question 7

$$12 \times 10^{-10} N$$

$$1.2 \times 10^{-10} N$$

$$0.12 \times 10^{-10} N$$

0.012 N

Solution

Question 8

$$v \propto r$$

$$\displaystyle v \propto \dfrac{1}{r}$$

$$\displaystyle v \propto \dfrac{1}{\sqrt{2}}r$$

$$v$$ is independent of r

Solution

Question 9

$$12\ F$$

$$4\ F$$

$$8\ F$$

$$16\ F$$

Solution

Question 10

$$100$$

$$141.4$$

$$173.2$$

$$200$$

Solution

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