Electric Charges and Fields Test

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Electric Charges and Fields Test - 72

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Weekly Quiz Competition

Question 1
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Two point charges $$q$$ and $$-4q$$ are placed at points $$A$$ and $$B$$ respectively, at a distance $$d$$ apart as shown in the figure. The null point (the point where electric field is zero) is :

A
$$ \dfrac {d}{3} $$ to the right of $$A$$

B
$$ \dfrac {d}{3} $$ to the left of $$B$$

C
$$ d $$ to the left of $$A$$

D
$$ d $$ to the right of $$B$$

Solution

Let the third charge is placed at distance\[x\] which is the null point.

So by using Coulombs law we get

$$\dfrac{{Kq}}{{{x^2}}} = \dfrac{{K4q}}{{{{\left( {d - x} \right)}^2}}}$$

By solving the above equation we get

$$d - x = 2x$$

Hence the distance of the third charge is$$x = \dfrac{d}{3}$$

The magnitude can be determined by the equilibrium condition. The charge on the position$$B$$will exert a force towards itself and the charge on position$$A$$will exert force away from it. Hence the third charge should be placed in between these two charges.
Question 2
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A cubical region of side $$'a'$$ has its centre at the origin. It encloses three fixed point charges, $$-q$$ at $$(0, -a/4, 0), +3q$$ at $$(0, 0, 0)$$ and $$-q$$ at $$(0, +a/4, 0)$$. Choose the incorrect option

A
The net electric flux crossing the plane $$x = + \dfrac {a}{2}$$ is equal to the net electric flux crossing the plane $$x = -\dfrac {a}{2}$$

B
The net electric flux crossing the plane $$y = +\dfrac {a}{2}$$ is more than the net electric flux crossing the plane $$y = -\dfrac {a}{2}$$

C
The net electric flux crossing the entire region is $$\dfrac {q}{\epsilon_{0}}$$

D
The net electric flux crossing the plane $$z = \dfrac {a}{2}$$ is equal to the net electric flux crossing the plane $$x = + \dfrac {a}{2}$$

Solution
Question 3
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Two equal and opposite point charges are kept in a box. A vacuum is created in the box by evacuating the air inside the box. Then the electrostatic force between the two charges will:

A
Increase

B
Decrease

C
Be zero

D
Be infinite

Solution

The electrostatic force between two charge are given as

$$F = \dfrac{{Qq}}{{4\pi {\varepsilon _0}{r^2}}}$$

The electrostatic force is inversely proportional to that of the dielectric permittivity.

The permittivity of free space (vacuum) is 1 and hence the electrostatic field in vacuum is more as there is no resistance to the field.
Question 4
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A particle of charges $$q_{0}$$ is moved around a charges $$+q$$ along the semicircle path of radius $$r$$ from $$A$$ to $$B$$ (see figure). The work done by the Coulomb force is

A
$$\dfrac {1}{4 \pi \epsilon _{0}}\dfrac {qq_{0}}{r}$$

B
$$\dfrac {1}{4 \pi \epsilon _{0}}\dfrac {2qq_{0}}{r}$$

C
$$\dfrac {1}{4 \pi \epsilon _{0}}\dfrac {qq_{0}}{r^{2}}\pi r$$

D
$$Zero$$

Solution
Question 5
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A charged particle $$q$$ is placed at the centre $$O$$ of cube of side $$L (ABCDEFGH)$$. Another same charge $$q$$ is placed at a distance $$L$$ from $$O$$, then the electric flux through $$ABCD$$ is

A
$$\dfrac {q}{4\pi \epsilon_{0}L}$$

B
Zero

C
$$\dfrac {q}{2\pi \epsilon_{0}L}$$

D
$$\dfrac {q}{3\pi \epsilon_{0}L}$$

Solution
Question 6
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Two charges $$q_{1}$$ and $$q_{2}$$ are moving with velocity $$\vec {V_{1}}$$ and $$\vec {V_{2}}$$ parallel to each other in vacuum as shown. Then,

A
Magnetic force between them is zero if $$\vec {V_{1}}$$ and $$\vec {V_{2}}$$ are equal at this instant

B
The ratio of magnitudes of magnetic force to electric is $$\dfrac {V_{1}V_{2}}{C^{2}}$$

C
The ratio of magnitude of magnetic force to electric is $$\sqrt {\dfrac{V_1 V_2}{C}}$$

D
The electric force between them is zero if $$\left| { \vec { V } }_{ 1 } \right| =\left| { \vec { V } }_{ 2 } \right|$$

Solution
Question 7
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Two point charges $$+q$$ and $$-4q$$ are placed at $$(-a, 0)$$ and $$(+a, 0)$$. Take electric field intensity to be positive if it is along positive $$x-$$direction. The variation of the electric field intensity as one moves along the $$x-$$axis is:

A

B

C

D

Solution
Question 8
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To construct an air filled capacitor which can store $$ 12 \mu C $$ of charge, What can be the minimum plate area of the capacitor? (Dielectric strength of air is $$3\times 10^6 V/m$$ )

A
$$0.25 m^2$$

B
$$0.45 m^2$$

C
$$4.5 m^2$$

D
$$0.35 m^2$$

Solution
Question 9
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Three charge $$+4q,\ Q$$ and $$q$$ are placed in a straight line of length $$\ell$$ at points distance $$0,\ \ell/2$$ and $$\ell$$ 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
$$\dfrac{ -q }{ 2 }$$

D
$$4q$$

Solution

Two forces are acting on $$q$$, one due to $$4q$$ and second due to $$Q$$.
$${F}_{1}=\dfrac{1}{4\pi\epsilon}\left(4q\right)\dfrac{q}{l^2}$$ and
$${F}_{2}=\dfrac{1}{4\pi\epsilon}\left(Q\right)\dfrac{q}{l^2/4}$$
And $${F}_{1}+{F}_{2}=0$$
$$\Rightarrow \dfrac{1}{4\pi\epsilon}\left(4q\right)\dfrac{q}{l^2}+\dfrac{1}{4\pi\epsilon}\left(4Q\right)\dfrac{q}{l^2}=0$$
$$\Rightarrow \dfrac{1}{4\pi\epsilon}\left(\left(4q\right)+\left(4Q\right)\right)\dfrac{q}{1^2}=0$$
$$\Rightarrow \dfrac{1}{\pi\epsilon}\left(\left(q\right)+\left(Q\right)\right)\dfrac{q}{1^2}=0$$
$$\therefore Q=-q$$
Question 10
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Consider a hemispherical surface of radius $$r$$, a positive point charge $$q$$ is kept at the centre of hemisphere. The electric flux through this hemisphere is :

A
zero

B
$$\dfrac{q}{\varepsilon_0}$$

C
$$\dfrac{q}{2\varepsilon_0}$$

D
$$\dfrac{2q}{\varepsilon_0}$$

Solution

if this charge is put at the centre of a sphere then net flux
$$=\phi =q/\varepsilon_0$$ ( hauss s law)
$$\Rightarrow$$ But this is hemisphere and it contain only half of the field lines and halve flux will become halved
Hence net flux through hemisphere $$=\dfrac{q}{2\varepsilon_0}$$