Electrostatics — Page 7 | JEE Physics

Q61

A uniformly charged disc of radius R having surface charge density

\sigma

is placed in the xy plane with its center at the origin. Find the electric field intensity along the z-axis at a distance Z from origin :-

A

E = {\sigma \over {2{\varepsilon _0}}}\left( {1 - {Z \over {{{({Z^2} + {R^2})}^{1/2}}}}} \right)

B

E = {\sigma \over {2{\varepsilon _0}}}\left( {1 + {Z \over {{{({Z^2} + {R^2})}^{1/2}}}}} \right)

C

E = {{2{\varepsilon _0}} \over \sigma }\left( {{1 \over {{{({Z^2} + {R^2})}^{1/2}}}} + Z} \right)

D

E = {\sigma \over {2{\varepsilon _0}}}\left( {{1 \over {({Z^2} + {R^2})}} + {1 \over {{Z^2}}}} \right)

Correct Answer

Option A

Solution

Consider a small ring of radius r and thickness dr on disc. area of elemental ring on disc dA = 2 $\pi$ rdr charge on this ring dq = $\sigma$ dA

dEz = {{kdqz} \over {{{({z^2} + {r^2})}^{3/2}}}}

E = \int\limits_0^R {d{E_z} = {\sigma \over {2{ \in _0}}}\left[ {1 - {z \over {\sqrt {{R^2} + {z^2}} }}} \right]}

Q62

Given below two statements : One is labelled as Assertion (A) and other is labelled as Reason (R). Assertion (A) : Non-polar materials do not have any permanent dipole moment. Reason (R) : When a non-polar material is placed in an electric field, the centre of the positive charge distribution of it's individual atom or molecule coincides with the centre of the negative charge distribution. In the light of above statements, choose the most appropriate answer from the options given below.

A Both (A) and (R) are correct and (R) is the correct explanation of (A).

B Both (A) and (R) are correct and (R) is not the correct explanation of (A).

C (A) is correct but (R) is not correct.

D (A) is not correct but (R) is correct.

Correct Answer

Option C

Solution

Non-polar bonds do not have any net dipole moment and are generally formed in compound where there is presence of symmetry.

When non polar material placed in electric field, due to redistribution of charges dipole is formed.

So, (R) is incorrect.

Q63

Sixty four conducting drops each of radius 0.02 m and each carrying a charge of 5

\mu

C are combined to form a bigger drop. The ratio of surface density of bigger drop to the smaller drop will be :

A 1 : 4

B 4 : 1

C 1 : 8

D 8 : 1

Correct Answer

Option B

Solution

Surface charge density of a spherical conductor is given by,

\sigma=\frac{q}{4 \pi r^2}

When all smaller drops combine, the radius of the bigger drop is given by $\dfrac{4}{3} \pi R^3=64\left(\dfrac{4}{3} \pi r^3\right)$

\begin{aligned} \mathrm{R}^3 & =64 r^3 \\\\ \mathrm{R} & =4 r \\\\ \sigma^{\prime} & =\frac{64 q}{4 \pi(4 r)^2}=4 \sigma \end{aligned}

\text{ Hence, } \frac{\sigma^{\prime}}{\sigma}=4: 1

Q64

Two identical charged particles each having a mass 10 g and charge 2.0

\times

10

-

7C are placed on a horizontal table with a separation of L between them such that they stay in limited equilibrium. If the coefficient of friction between each particle and the table is 0.25, find the value of L. [Use g = 10 ms

-

2]

A 12 cm

B 10 cm

C 8 cm

D 5 cm

Correct Answer

Option A

Solution

According to given information:

{{k{Q^2}} \over {{L^2}}}

= $\mu$ mg Putting the values, we get L = 12 cm

Q65

A long cylindrical volume contains a uniformly distributed charge of density

\rho

. The radius of cylindrical volume is R. A charge particle (q) revolves around the cylinder in a circular path. The kinetic energy of the particle is :

A

{{\rho q{R^2}} \over {4{\varepsilon _0}}}

B

{{\rho q{R^2}} \over {2{\varepsilon _0}}}

C

{{q\rho } \over {4{\varepsilon _0}{R^2}}}

D

{{4{\varepsilon _0}{R^2}} \over {q\rho }}

Correct Answer

Option A

Solution

{{m{v^2}} \over r} = {{2k\rho \times \pi {R^2}q} \over r}

\Rightarrow {1 \over 2}m{v^2} = {{\rho {R^2}q} \over {4{\varepsilon _0}}}

Q66

Two uniformly charged spherical conductors

A

and

B

of radii

5 \mathrm{~mm}

and

10 \mathrm{~mm}

are separated by a distance of

2 \mathrm{~cm}

. If the spheres are connected by a conducting wire, then in equilibrium condition, the ratio of the magnitudes of the electric fields at the surface of the sphere

A

and

B

will be :

A 1 : 2

B 2 : 1

C 1 : 1

D 1 : 4

Correct Answer

Option B

Solution

After connection

{\sigma _1}{R_1} = {\sigma _2}{R_2}

Now

E = {\sigma \over {{\varepsilon _0}}}

\Rightarrow {{{E_1}} \over {{E_2}}} = {{{\sigma _1}} \over {{\sigma _2}}} = {{{R_2}} \over {{R_1}}} = {2 \over 1}

Q67

Two identical positive charges

Q

each are fixed at a distance of '2a' apart from each other. Another point charge

q_{0}

with mass 'm' is placed at midpoint between two fixed charges. For a small displacement along the line joining the fixed charges, the charge

\mathrm{q}_{0}

executes

\mathrm{SHM}

. The time period of oscillation of charge

\mathrm{q}_{0}

will be :

A

\sqrt{\dfrac{4 \pi^{3} \varepsilon_{0} m a^{3}}{q_{0} Q}}

B

\sqrt{\dfrac{q_{0} Q}{4 \pi^{3} \varepsilon_{0} m a^{3}}}

C

\sqrt{\dfrac{2 \pi^{2} \varepsilon_{0} m a^{3}}{q_{0} Q}}

D

\sqrt{\dfrac{8 \pi^{3} \varepsilon_{0} m a^{3}}{q_{0} Q}}

Correct Answer

Option A

Solution

(x << a) ( $\alpha$ is acceleration)

{F_{net}} = - \left( {{{k{q_0}Q} \over {{{(a - x)}^2}}} - {{kQ{q_0}} \over {{{(a + x)}^2}}}} \right)

m\alpha = - {{k{q_0}Q} \over {{a^4}}}4ax

\Rightarrow \alpha = - {{4k{q_0}Q} \over {m{a^3}}}x

So,

T = 2\pi \sqrt {{{4\pi {\varepsilon _0}m{a^3}} \over {4{q_0}Q}}}

or

T = \sqrt {{{4{\pi ^3}{\varepsilon _0}m{a^3}} \over {{q_0}Q}}}

Q68

A charge of

4 \,\mu \mathrm{C}

is to be divided into two. The distance between the two divided charges is constant. The magnitude of the divided charges so that the force between them is maximum, will be :

A

1 \,\mu \mathrm{C}

and

3 \,\mu\mathrm{C}

B

2 \,\mu \mathrm{C}

and

2\, \mu \mathrm{C}

C 0 and

4\, \mu\, \mathrm{C}

D

1.5 \,\mu \mathrm{C}

and

2.5\, \mu \mathrm{C}

Correct Answer

Option B

Solution

so

F = {{kq(4 - q) \times {{10}^{ - 12}}} \over {{r^2}}}

so Fmax will be at q = 2 $\mu$ C

Q69

A point charge of 10

\mu

C is placed at the origin. At what location on the X-axis should a point charge of 40

\mu

C be placed so that the net electric field is zero at

x=2

cm on the X-axis?

A

x=6

cm

B

x=8

cm

C

x=4

cm

D

x=-4

cm

Correct Answer

Option A

Solution

\therefore E_{x}=2 \mathrm{~cm}=0

\begin{aligned} & \frac{1}{4 \pi \varepsilon_{0}} \frac{(10 \mu \mathrm{C})}{(2 \mathrm{~cm})^{2}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{(40 \mu \mathrm{C})}{[(\mathrm{a}-2) \mathrm{cm}]^{2}} \end{aligned}

$\Rightarrow\left(\dfrac{a-2}{2}\right)^{2}=4$ $\Rightarrow \dfrac{a-2}{2}=2$

a=6 \mathrm{~cm}

Q70

Given below are two statements. Statement I : Electric potential is constant within and at the surface of each conductor. Statement II : Electric field just outside a charged conductor is perpendicular to the surface of the conductor at every point. In the light of the above statements, choose the most appropriate answer from the options given below.

A Both Statement I and Statement II are correct

B Both Statement I and Statement II are incorrect

C Statement I is correct but Statement II is incorrect

D Statement I is incorrect but Statement II is correct

Correct Answer

Option A

Solution

Since

{\overrightarrow E _{net}} = \overrightarrow 0

in the bulk of a conductor $\Rightarrow$ Potential would be constant.

$\Rightarrow$ Statement I is correct Since a conductor's surface is equipotential,

\overrightarrow E

just outside is perpendicular to the surface.