Electrostatics — Page 10 | NEET Physics

Q91

Electric field in a certain region is given by

\overrightarrow{\mathrm{E}}=\left(\dfrac{\mathrm{A}}{x^{2}} \hat{i}+\dfrac{\mathrm{B}}{y^{3}} \hat{j}\right) \text{. The } \mathrm{SI} \text{ unit of } \mathrm{A} \text{ and } \mathrm{B}

are :

A

\mathrm{Nm}^{2} \mathrm{C} ; \mathrm{Nm}^{3} \mathrm{C}

B

\mathrm{Nm}^{3} \mathrm{C}^{-1} ; \mathrm{Nm}^{2} \mathrm{C}^{-1}

C

\mathrm{Nm}^{3} \mathrm{C} ; \mathrm{Nm}^{2} \mathrm{C}

D

\mathrm{Nm}^{2} \mathrm{C}^{-1} ; \mathrm{Nm}^{3} \mathrm{C}^{-1}

Correct Answer

Option D

Solution

\overrightarrow E = \left( {{A \over {{x^2}}}\widehat i + {B \over {{y^3}}}\widehat j} \right)

\left[ {{A \over {{x^2}}}} \right] = [E] = \left[ {{F \over q}} \right] = \left[ {{N \over C}} \right] = N{C^{ - 1}}

\mathrm{[A] = (N{m^2}{C^{ - 1}})}

\mathrm{[B] = N{m^3}{C^{ - 1}}}

Q92

In a metallic conductor, under the effect of applied electric field, the free electrons of the conductor

A move in the straight line paths in the same direction

B move with the uniform velocity throughout from lower potential to higher potential

C drift from higher potential to lower potential.

D move in the curved paths from lower potential to higher potential

Correct Answer

Option D

Solution

Electron drifts from lower potential to higher potential on curved path.

Q93

Two charges

q

and

3 q

are separated by a distance '

r

' in air. At a distance

x

from charge

q

, the resultant electric field is zero. The value of

x

is :

A

\dfrac{r}{3(1+\sqrt{3})}

B

\dfrac{(1+\sqrt{3})}{r}

C

\dfrac{r}{(1+\sqrt{3})}

D

r(1+\sqrt{3})

Correct Answer

Option C

Solution

\begin{aligned} & \left(\vec{E}_{\text{net }}\right)_P=0 \\ & \frac{\mathrm{kq}}{\mathrm{x}^2}=\frac{\mathrm{k} \cdot 3 \mathrm{q}}{(\mathrm{r}-\mathrm{x})^2} \\ & (\mathrm{r}-\mathrm{x})^2=3 \mathrm{x}^2 \\ & \mathrm{r}-\mathrm{x}=\sqrt{3} \mathrm{x} \\ & \mathrm{x}=\frac{\mathrm{r}}{\sqrt{3}+1} \end{aligned}

Q94

Given below are two statements: Statement I : An electric dipole is placed at the center of a hollow sphere. The flux of the electric field through the sphere is zero but the electric field is not zero anywhere in the sphere. Statement II : If R is the radius of a solid metallic sphere and Q be the total charge on it. The electric field at any point on the spherical surface of radius r (

A Both Statement I and Statement II are true

B Statement I is false but Statement II is true

C Statement I is true but Statement II is false

D Both Statement I and Statement II are false

Correct Answer

Option C

Solution

Net charge on electric dipole = + q $-$ q = 0 Hence, according to Gauss's law, Electric flux,

\phi = {{{q_{net}}} \over {{\varepsilon _0}}} = {0 \over {{\varepsilon _0}}} = 0

Electric field due to electric dipole is non-zero and varies at point to point.

Hence, statement I is true.

Electric field due to charged solid sphere at a distance r from centre.

E = {1 \over {4\pi {\varepsilon _0}}}\,.\,{{Qr} \over {{R^3}}}

[when r Hence, statement II is false. Hence, option (c) is the correct.

Q95

A

10 ~\mu \mathrm{C}

charge is divided into two parts and placed at

1 \mathrm{~cm}

distance so that the repulsive force between them is maximum. The charges of the two parts are:

A

9 ~\mu\mathrm{C}, 1 ~\mu \mathrm{C}

B

5 ~\mu\mathrm{C}, 5 ~\mu \mathrm{C}

C

8 ~\mu\mathrm{C}, 2 ~\mu \mathrm{C}

D

7 ~\mu\mathrm{C}, 3 ~\mu \mathrm{C}

Correct Answer

Option B

Solution

The repulsive force between the two charges is given by Coulomb's law:

F=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2},

where $F$ is the force, $q_1$ and $q_2$ are the charges, $r$ is the distance between them, and $\epsilon_0$ is the electric constant.

To maximize the force, we need to maximize the product $q_1q_2$ .

Let $q_1$ be the charge on one part and $q_2$ be the charge on the other part.

Then we have $q_1+q_2=10\mu\mathrm{C}$ , since the total charge is $10\mu\mathrm{C}$ .

The distance between the two charges is $r=1\mathrm{~cm}=0.01\mathrm{~m}$ .

To maximize the force, we need to maximize $q_1q_2$ , subject to the constraint that $q_1+q_2=10\mu\mathrm{C}$ .

We can use the method of Lagrange multipliers to find the values of $q_1$ and $q_2$ that maximize $q_1q_2$ subject to the constraint $q_1+q_2=10\mu\mathrm{C}$ .

The Lagrangian is given by

\mathcal{L}=q_1q_2-\lambda(q_1+q_2-10\mu\mathrm{C}),

where $\lambda$ is the Lagrange multiplier.

Taking the partial derivatives of $\mathcal{L}$ with respect to $q_1$ , $q_2$ , and $\lambda$ , and setting them equal to zero, we get:

q_2-\lambda=0

q_1-\lambda=0

q_1+q_2=10\mu\mathrm{C}

Solving for $q_1$ and $q_2$ , we get $q_1=q_2=5\mu\mathrm{C}$ .

Therefore, the charges of the two parts are both $5\mu\mathrm{C}$ .

Therefore, to maximize the repulsive force between the two charges, we need to divide the $10\mu\mathrm{C}$ charge into two equal parts of $5\mu\mathrm{C}$ each, and place them $1\mathrm{~cm}$ apart.

Q96

A point charge causes an electric flux of

-2 \times 10^4 \mathrm{Nm}^2 \mathrm{C}^{-1}

to pass through a spherical Gaussian surface of 8.0 cm radius, centred on the charge. The value of the point charge is : (Given

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

)

A

17.7 \times 10^{-8} \mathrm{C}

B

-17.7 \times 10^{-8} \mathrm{C}

C

15.7 \times 10^{-8} \mathrm{C}

D

-15.7 \times 10^{-8} \mathrm{C}

Correct Answer

Option B

Solution

Given an electric flux $\phi = -2 \times 10^4 \, \text{Nm}^2\text{C}^{-1}$ through a spherical Gaussian surface with a radius of 8.0 cm (r = 0.08 m), we are to find the value of the point charge $Q$ .

According to Gauss's Law: $\phi = \dfrac{Q}{\varepsilon_0}$ Where $\varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2\text{N}^{-1}\text{m}^{-2}$ .

Rearranging the equation to solve for $Q$ , we have: $Q = \phi \varepsilon_0$ Substituting the given values: $Q = -2 \times 10^4 \times 8.85 \times 10^{-12}$ Calculating this yields: $Q = -17.7 \times 10^{-8} \, \text{C}$ Thus, the value of the point charge is $Q = -17.7 \times 10^{-8} \, \text{C}$ .

Q97

A positive charge particle of 100 mg is thrown in opposite direction to a uniform electric field of strength 1

\times

105 NC

-

1. If the charge on the particle is 40

\mu

C and the initial velocity is 200 ms

-

1, how much distance it will travel before coming to the rest momentarily :

A 1 m

B 5 m

C 10 m

D 0.5 m

Correct Answer

Option D

Solution

{v^2} - {u^2} = 2as

\Rightarrow {0^2} - {200^2} = 2\left( {{{ - qE} \over m}} \right)(S)

\Rightarrow - {200^2} = 2\left[ {{{ - 40 \times {{10}^{ - 6}} \times {{10}^5}} \over {100 \times {{10}^{ - 6}}}}} \right][S]

\Rightarrow S = {4 \over {2 \times 4}}

m = 0.5 m

Q98

The potential (in volts) of a charge distribution is given by. V(z) = 30

-

5x2 for

\left| z \right|

\le

1 m. V(z) = 35

-

10

\left| z \right|

for

\left| z \right|

\ge

1 m. V(z) does not depend on x and y. If this potential is generated by a constant charge per unit volume

{\rho _0}

(in units of

{\varepsilon _0}

) which is spread over a certain region, then choose the correct statement.

A

{\rho _0}

= 10

{\varepsilon _0}

for

\left| z \right|

\le

1 m and

{\rho _0} = 0

elsewhere

B

{\rho _0}

= 20

{\varepsilon _0}

in the entire region

C

{\rho _0}

= 40

{\varepsilon _0}

in the entire region

D

{\rho _0}

= 20

{\varepsilon _0}

for

\left| z \right|

\le

1 m and

{\rho _0} = 0

elsewhere

Correct Answer

Option A

Solution

We know, E(z) = $-$

{{dv} \over {dz}}

$\therefore$ E(z) = $-$ 10 z for

\left| z \right| \le 1

m and E(z) = 10 for

\left| z \right| \ge 1

m $\therefore$ The source is an infinity large non conducting thick of thickness z = 2 m. $\therefore$ E =

{\sigma \over {2{\varepsilon _0}}}

=

{{\rho \left( 2 \right)} \over {2{\varepsilon _0}}}

=

{\rho \over {{\varepsilon _0}}}

$\therefore$

{\rho \over {{\varepsilon _0}}}

= 10 $\Rightarrow$ $\rho$ =

{10\,{\varepsilon _0}}