Electrostatics — Page 9 | NEET Physics

Q81

Two charges

7 \mu \mathrm{c}

and

-4 \mu \mathrm{c}

are placed at

(-7 \mathrm{~cm}, 0,0)

and

(7 \mathrm{~cm}, 0,0)

respectively. Given,

\epsilon_0=8.85 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}

, the electrostatic potential energy of the charge configuration is :

A

-2.0

J

B

-1.5

J

C

-1.2

J

D

-1.8

J

Correct Answer

Option D

Solution

U = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r}

Here,

q_1 = 7 \times 10^{-6} \, C

q_2 = -4 \times 10^{-6} \, C

The separation between the charges is along the x-axis from

-7 \, \text{cm}

to

7 \, \text{cm}

, so

r = 0.14 \, m

The Coulomb constant is approximated as

\frac{1}{4\pi\epsilon_0} \approx 9 \times 10^9 \, \frac{Nm^2}{C^2}

Substitute these values into the equation:

U = 9 \times 10^9 \times \frac{(7 \times 10^{-6})(-4 \times 10^{-6})}{0.14}

Calculate the product of the charges:

(7 \times 10^{-6})(-4 \times 10^{-6}) = -28 \times 10^{-12} \, C^2

Now, divide by the distance:

\frac{-28 \times 10^{-12}}{0.14} = -2 \times 10^{-10} \, C^2/m

Finally, multiply by the Coulomb constant:

U = 9 \times 10^9 \times (-2 \times 10^{-10}) = -1.8 \, J

Thus, the electrostatic potential energy of the configuration is:

-1.8 \, J

This corresponds to Option D.

Q82

Two small spherical balls of mass 10 g each with charges

-2 \mu \mathrm{C}

and

2 \mu \mathrm{C}

, are attached to two ends of very light rigid rod of length 20 cm . The arrangement is now placed near an infinite nonconducting charge sheet with uniform charge density of

100 \mu \mathrm{C} / \mathrm{m}^2

such that length of rod makes an angle of

30^{\circ}

with electric field generated by charge sheet. Net torque acting on the rod is: (Take

\varepsilon_{\mathrm{o}}: 8.85 \times 10^{-12} \mathrm{C}^2 / \mathrm{Nm}^2

)

A 1.12 Nm

B 2.24 Nm

C 11.2 Nm

D 112 Nm

Correct Answer

Option A

Solution

\begin{aligned} & \mathrm{E}=\frac{\sigma}{2 \varepsilon_0} \\ & \tau=\mathrm{PE} \sin \theta \\ & =\left[\left(2 \times 10^{-6}\right)\left(\frac{2}{10}\right)\right]\left[\frac{100 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}}\right]\left(\frac{1}{2}\right) \\ & =\frac{10}{8.85}=1.12 \mathrm{Nm} \end{aligned}

Q83

Two charges of

5 Q

and

-2 Q

are situated at the points

(3 a, 0)

and

(-5 a, 0)

respectively. The electric flux through a sphere of radius '

4 a

' having center at origin is :

A

\dfrac{2 Q}{\varepsilon_0}

B

\dfrac{7 Q}{\varepsilon_0}

C

\dfrac{3 Q}{\varepsilon_0}

D

\dfrac{5 Q}{\varepsilon_0}

Correct Answer

Option D

Solution

The electric flux through any closed surface is given by Gauss's law, which can be stated as:

\Phi = \frac{Q_{\text{enc}}}{\varepsilon_0}

where:

\Phi

= electric flux

Q_{\text{enc}}

= total charge enclosed by the surface

\varepsilon_0

= permittivity of free space (electric constant) In this scenario, we have a sphere of radius

4a

with its center at the origin and two charges,

5Q

at point (3a, 0) and

-2Q

at point (-5a, 0). Since the sphere's radius is

4a

, the charge

5Q

, which is located at (3a, 0), lies inside the sphere, whereas the charge

-2Q

, located at (-5a, 0), lies outside the sphere.

Gauss's law only considers charges that are enclosed within the surface.

Thus, the charge

-2Q

has no impact on the electric flux through the sphere because it is not enclosed by the sphere. Only the charge

5Q

contributes to the electric flux inside the sphere.

The electric flux through the sphere is therefore given simply by the enclosed charge:

\Phi = \frac{5Q}{\varepsilon_0}

Q84

If

\epsilon_0

denotes the permittivity of free space and

\Phi_E

is the flux of the electric field through the area bounded by the closed surface, then dimensions of

\left(\epsilon_0 \dfrac{d \phi_E}{d t}\right)

are that of :

A electric charge

B electric field

C electric current

D electric potential

Correct Answer

Option C

Solution

We can use Gauss’s law to analyze the dimensions. Gauss’s law states that

\Phi_E = \frac{q}{\epsilon_0},

where

\Phi_E

is the electric flux,

q

is the electric charge,

\epsilon_0

is the permittivity of free space. To find the dimensions of

\epsilon_0 \frac{d\Phi_E}{dt}

, follow these steps: Differentiate Gauss’s law with respect to time:

\frac{d\Phi_E}{dt} = \frac{1}{\epsilon_0}\frac{dq}{dt}.

Multiply the equation by

\epsilon_0

:

\epsilon_0 \frac{d\Phi_E}{dt} = \frac{dq}{dt}.

Recognize that the time rate of change of charge,

\frac{dq}{dt}

, is by definition the electric current. Thus, the dimensions of

\epsilon_0 \frac{d\Phi_E}{dt}

are those of electric current. The correct answer is: Option C, electric current.

Q85

Two electrons each are fixed at a distance '2d'. A third charge proton placed at the midpoint is displaced slightly by a distance x (x << d) perpendicular to the line joining the two fixed charges. Proton will execute simple harmonic motion having angular frequency : (m = mass of charged particle)

A

{\left( {{{2{q^2}} \over {\pi {\varepsilon _0}m{d^3}}}} \right)^{{1 \over 2}}}

B

{\left( {{{{q^2}} \over {2\pi {\varepsilon _0}m{d^3}}}} \right)^{{1 \over 2}}}

C

{\left( {{{2\pi {\varepsilon _0}m{d^3}} \over {{q^2}}}} \right)^{{1 \over 2}}}

D

{\left( {{{\pi {\varepsilon _0}m{d^3}} \over {2{q^2}}}} \right)^{{1 \over 2}}}

Correct Answer

Option B

Solution

The arrangement of charges is shown below As we know that, Coulomb's force between two charges. i.e., q1 and q2,

F = {1 \over {4\pi {\varepsilon _0}}}{{{q_1}{q_2}} \over {{r^2}}} = {1 \over {4\pi {\varepsilon _0}}}{{{q_1}{q_2}} \over {({d^2} + {x^2})}}

..... (i) Here,

{q_1} = {q_2} = q

Force in SHM,

F = m{\omega ^2}x

...... (ii) Since, in order to have SHM +q should move downwards and force responsible for this will be only

F' = F\sin \theta + F\sin \theta = 2F\sin \theta

..... (iii) Using Eqs. (ii) and (iii), we get

2F\sin \theta = m{\omega ^2}x

\Rightarrow {2 \over {4\pi {\varepsilon _0}}}{{{q^2}} \over {({d^2} + {x^2})}}\sin \theta = m{\omega ^2}x

\Rightarrow {2 \over {4\pi {\varepsilon _0}}}{{{q^2}} \over {({d^2} + {x^2})}}.{x \over {{{({d^2} + {x^2})}^{1/2}}}} = m{\omega ^2}x

\Rightarrow \omega = {\left( {{1 \over {2\pi {\varepsilon _0}}}{{{q^2}} \over {{{({d^2} + {x^2})}^{3/2}}m}}} \right)^{1/2}}

As,

x < < d

$\therefore$

\omega = {\left( {{1 \over {2\pi {\varepsilon _0}}}{{{q^2}} \over {m{d^3}}}} \right)^{1/2}}

Q86

In hydrogen like system the ratio of coulombian force and gravitational force between an electron and a proton is in the order of :

A 1019

B 1039

C 1029

D 1036

Correct Answer

Option B

Solution

To find the ratio of the Coulombic force to the gravitational force between an electron and a proton in a hydrogen-like system, we use the formulae for both forces and then divide them.

The Coulombic (electrostatic) force, $F_C$ , between two charges is given by Coulomb's law: $F_C = k \dfrac{|q_1 q_2|}{r^2}$ where $k$ is Coulomb's constant ( $8.987 \times 10^9 \, \text{Nm}^2\text{C}^{-2}$ ), $q_1$ and $q_2$ are the magnitudes of the charges, and $r$ is the distance between the charges.

For a proton and an electron, $q_1 = q_2 = e$ , where $e$ is the elementary charge ( $1.602 \times 10^{-19} \, \text{C}$ ).

The gravitational force, $F_G$ , between two masses is given by Newton's law of universal gravitation: $F_G = G \dfrac{m_1 m_2}{r^2}$ where $G$ is the gravitational constant ( $6.674 \times 10^{-11} \, \text{Nm}^2\text{kg}^{-2}$ ), and $m_1$ and $m_2$ are the masses of the two objects.

For a proton and an electron, $m_p \approx 1.673 \times 10^{-27} \, \text{kg}$ and $m_e \approx 9.109 \times 10^{-31} \, \text{kg}$ , respectively.

The ratio of the Coulombic to the gravitational force is therefore: $\dfrac{F_C}{F_G} = \dfrac{k \dfrac{|e^2|}{r^2}}{G \dfrac{m_p m_e}{r^2}} = \dfrac{k e^2}{G m_p m_e}$ Plugging in the values: $\dfrac{F_C}{F_G} = \dfrac{(8.987 \times 10^9) (1.602 \times 10^{-19})^2}{(6.674 \times 10^{-11}) (1.673 \times 10^{-27}) (9.109 \times 10^{-31})}$ $\dfrac{F_C}{F_G} = \dfrac{(8.987 \times 10^9) \times (2.568 \times 10^{-38})}{(6.674 \times 10^{-11}) \times (1.523 \times 10^{-57})}$ $\dfrac{F_C}{F_G} \approx 2.3 \times 10^{39}$ Therefore, the ratio of the Coulombic force to the gravitational force between an electron and a proton in a hydrogen-like system is on the order of $10^{39}$ , making Option B the correct answer.

Q87

A vertical electric field of magnitude 4.9

\times

105 N/C just prevents a water droplet of a mass 0.1 g from falling. The value of charge on the droplet will be : (Given : g = 9.8 m/s2)

A 1.6

\times

10

-

9 C

B 2.0

\times

10

-

9 C

C 3.2

\times

10

-

9 C

D 0.5

\times

10

-

9 C

Correct Answer

Option B

Solution

Since the droplet is at rest $\Rightarrow$ Net force = 0 $\Rightarrow$ mg = qE $\Rightarrow$

q = {{mg} \over E}

= 2 $\times$ 10 $-$ 9 C

Q88

Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle of

{30^ \circ }

with each other. When suspended in a liquid of density

0.8g

c{m^{ - 3}},

the angle remains the same. If density of the material of the sphere is

1.6

g

c{m^{ - 3}},

the dielectric constant of the liquid is

A

4

B

3

C

2

D

1

Correct Answer

Option C

Solution

{F_e} = T\sin {15^ \circ }\,\,;

mg = T\cos {15^ \circ }

\Rightarrow \tan {15^ \circ } = {{{F_e}} \over {mg}}

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)

In liquid,

{F_e}' = T'\sin {15^ \circ }

\,\,\,\,\,\,...(ii)

mg = {F_B} + T'\cos {15^ \circ }

{F_B}' = V\left( {d - \rho } \right)g = V\left( {1.6 - 0.8} \right)g = 0.8\,Vg

= 0.8{m \over d}g = {{0.8mg} \over {1.6}} = {{mg} \over 2}

$\therefore$

mg = {{mg} \over 2} + T'\cos {15^ \circ }

\Rightarrow {{mg} \over 2} = T'\cos {15^ \circ }

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( B \right)

From

(A)

and

(B),

\tan \,{15^ \circ } = {{2{F_e}'} \over {mg}}\,\,\,\,\,\,\,\,\,...\left( 2 \right)

From

(1)

and

(2)

{{{F_e}} \over {mg}} = {{2{F_e}'} \over {mg}} \Rightarrow {F_e} = 2{F_e}' \Rightarrow {F_e}' = {{{F_e}} \over 2}

Q89

For a uniformly charged ring of radius R, the electric field on its axis has the largest magnitude at a distance h from its center. Then value of h is :

A

{R \over {\sqrt 5 }}

B

{R \over {\sqrt 2 }}

C R

D R

\sqrt 2

Correct Answer

Option B

Solution

Electric field on the axis of the ring,

E = {{KQh} \over {{{\left( {{R^2} + {h^2}} \right)}^{{3 \over 2}}}}}

For maximum electric field,

{{dE} \over {dh}} = 0

$\Rightarrow$

h = {R \over {\sqrt 2 }}

Q90

Considering a group of positive charges, which of the following statements is correct ?

A Net potential of the system cannot be zero at a point but net electric field can be zero at that point

B Net potential of the system at a point can be zero but net electric field can't be zero at that point.

C Both the net potential and the net electric field cannot be zero at a point.

D Both the net potential and the net field can be zero at a point.

Correct Answer

Option A

Solution

$V=\dfrac{\sum K Q_{i}}{r_{i}}$ Here, $Q_{i}$ and $r_{i}$ are positive.

$\therefore V > 0$ The correct statement is: (A) Net potential of the system cannot be zero at a point but net electric field can be zero at that point.

Explanation: In a group of positive charges, the net potential at a point is the sum of the potentials due to each individual charge.

The potential due to a point charge is given by the Coulomb's law, which is non-zero except at the location of the charge itself.

Therefore, the net potential due to a group of positive charges can never be zero at a point.

On the other hand, the net electric field at a point is the vector sum of the electric fields due to each individual charge.

If the charges are arranged in such a way that their electric fields cancel out at a particular point, then the net electric field at that point can be zero, even though the charges are present.

This can happen, for example, in a symmetrical arrangement of charges.

So, statement (A) is the correct statement.