Physics 3 Marks Question

1. To the left of the plates, electric fields are equal and opposite as plates are close to each other electric field is zero as surface charge density in outer side is zero.
2. To the right of the plates, electric fields are equal and opposite as plates are close to each other electric field is zero.
3. Electric fields between the plates are in same direction as total E.F. on both sides of plate due to $\sigma$ surface charge density = $\frac{\sigma}{\in_0}$ electric field of inner side of plate =$\frac{\sigma}{2\in_0}$

and for both plate E $=\frac{\sigma}{2\in_0}+\frac{\sigma}{2\in_0}$

$\text{E}=\frac{\sigma}{\in_0}=\sigma\times4\pi\times9\times10^9$

E = 17.0 × 10^-22 × 4 × 3.14 × 9 × 10⁹

E = 1921.7 × 10^-13

= 1.92 × 10^-10 N/C.

1. Let us take a Gaussian surface which is lying completely within the conductor and enclosing the cavity. According to the Gaussian theorem, the charge enclosed by Gaussian surface must be zero as electric field vanishes everywhere inside a conductor. Thus, electric field vanishes inside the cavity. Therefore, charges which are supplied to the conductor reside on its outer surface.

2. Let us take a Gaussian surface inside the conductor which is quite close to the cavity. According to the Gaussian theorem,

$\phi_\text{E}=\int\text{E.ds}=\frac{\text{total charge}}{\in_0}$

As the electric field inside the conductor is zero, the total charge which is enclosed by the gaussian surface must be zero. This requires, a charge of -q units to be induced on the inner surface of the hollow conductor A. But an equal and opposite charge +q units must appear on the outer surface of conductor A, so that the total charge on the outer surface of A is Q + q.

3. Use a metallic surface to enclose the sensitive instrument completely safe and intact. Because of electrostatic shielding, the electric field inside the metal surface vanishes to zero and all charge will reside on outer surface.

1. Net outward flux through the surface of the box, ($\phi$) = 8.0 x 10³ Nm²/C

For a body containing net charge a, flux is given by the relation,

$\phi = \frac{\text{q}}{\in_0}$

$\in_0$ = Permittivity of free space

= 8.854 × 10^-12 × 8.0 × 10³

q = $\in_0\phi$

= 8.854 × 10^-12 × 8.0 × 10³

= 7.08 × 10^-8

= 0.07 µC

Therefore, the net charge inside the box is 0.07 µC.

2. No

Net flux piercing out through a body depends on the net charge contained in the body. If net flux is zero, then it can be inferred that net charge inside the body is zero. The body may have equal amount of positive and negative charges.

1. Electrostatic force on the first sphere, F = 0.2 N 

Charge on this sphere, q₁ = 0.4 µC = 0.4 × 10^-6 C

Charge on the second sphere, q₂ = -0.8 µC = 0.8 × 10^-6 C

Electrostatic force between the spheres is given by the relation,

$\text{F}=\frac{\text{q}_1\text{q}_2}{4\pi\in_0\text{r}^2}$ And $\frac{1}{4\pi\in_0}=9\times10^{9}\text{Nm}^2\text{C}^{-2}$

Where, $\in_0\ =$ Permittivity of free space

And, $\frac{1}{4\pi\in_0} = 9\times10^9\text{Nm}^{-2}\text{C}^{-2}$

$\text{r}^2=\frac{\text{q}_1\text{q}_2}{4\pi\in_0\text{F}}$

= 144 × 10^-4

r = $\sqrt{144\times10^{-4}}=0.12\text{m}$

The distance between the two spheres is 0.12m.

2. Both the spheres attract each other with the same force. Therefore, the force on the second sphere due to the first is 0.2N.

$\text{E }\delta\text{ S} =\frac{|\sigma| \delta\text{ S}}{\varepsilon_{0}}$

$\therefore\text{E} = \frac{|\sigma|}{\varepsilon_{0}}.$

1. Statement: Net electric flux through to a closed surface is equal to $\frac{1}{\varepsilon_\circ}$ times the total net charge enclosed within the surface.

(If the student just writes $\oint\text{E}.\text{ds} =\frac{\text{q}}{\in_\circ}$, award )

2. Diagram:- 

3. Derivation:- 

$\oint\text{E}.\text{ds} \int\limits_{s_1}\overline{E}.\text{d}\overline{s}_{1} + \int\limits_{s_2}\overline{E}.\text{ds}_{2} + \int\limits_{s_3}\overline{E}.\text{ds}_{3}$

$ = 0 + 0 + 2\pi\text{r}\ell$

Also, $\text{q} = \lambda$  $\ell$ (where $\lambda$  is charge per unit length)

$(\text{E}).(2\pi\text{r}\ell) = \frac{1}{\varepsilon_\circ}\lambda\ell $ OR $\text{E}2\pi\text{r}\ell\frac{\text{q}}{\varepsilon_\circ}$

$\text{E} = \frac{\lambda}{2\pi\varepsilon_\circ\text{r}}$ OR $\text{E} = \frac{\text{q}}{2\pi\varepsilon_\circ\text{r}\ell}$.

By Gauss’s law $ \oint \overrightarrow{E}.\overrightarrow{ds}=\frac{q}{\in_0}$

$\therefore{2{\text{EA}}}=\frac{\sigma{\text{A}}}{{\in_0}}$

$\therefore\text{E}=\frac{\sigma}{2\in_0}\text{ }\text{or}\frac{\sigma}{2\in_0}\text{A}$

$\because$  Both the sheets have same charge density, their electric fields will be equal and opposite in the region between the two sheets.

Alternate Answer

$\text{E}_1=\frac{\sigma}{2\in_0}$

$\text{E}_2=-\frac{\sigma}{2\in_0}$

$\frac{\sigma}{2\in_0}-\frac{\sigma}{2\in_0}=0$]

$\phi_E=\overrightarrow{E}.\overrightarrow{A}={EA}\cos\theta$

1. $\tau=pE\sin\theta\text{ };\text{ }\theta=$ angle between dipole moment($\overrightarrow{p}$) and electric field($\overrightarrow{\text{E}}$)

$\tau=\overrightarrow{p}\times\overrightarrow{\text{E}}$

Direction of torque is perpendicular to the plane containing and given by right-hand screw rule.

Alternate Answer

Direction of torque is out of the plane of the paper.

2. If the field is non uniform the net force on the dipole will not be zero. There will be translatory motion of the dipole.
3.  

1. Net force will be in the direction of increasing electric field.

2. Net force will be in the direction opposite to the increasing field. [or in the direction of decreasing field]

$\oint\overrightarrow{E}.\vec{ds}=\frac{q}{\in_0}$

1. $\oint\overrightarrow{E_1}.\vec{ds}=\frac{\lambda_1l}{\in_0}$

$\Longrightarrow\overrightarrow{E_1}=\frac{\lambda_1}{2\pi\in_0r_1}\hat{r_1}$

2. $\oint\overrightarrow{E_2}.\vec{ds}=\frac{(\lambda_1-\lambda_2)l}{\in_0}$

$\Longrightarrow\overrightarrow{E_2}=\frac{(\lambda_1-\lambda_2)}{2\pi\in_0r_2}\hat{r_2}$

1. $\tau=pE\sin\theta\text{ };\text{ }\theta=$ angle between dipole moment($\overrightarrow{p}$) and electric field($\overrightarrow{\text{E}}$)

$\tau=\overrightarrow{p}\times\overrightarrow{\text{E}}$

Direction of torque is perpendicular to the plane containing and given by right-hand screw rule.

Alternate Answer

Direction of torque is out of the plane of the paper.

2. If the field is non uniform the net force on the dipole will not be zero. There will be translatory motion of the dipole.
3.  

1. Net force will be in the direction of increasing electric field.

2. Net force will be in the direction opposite to the increasing field. [or in the direction of decreasing field]

Force on + q, $\overrightarrow{\text{F}}=q\overrightarrow{\text{E}}$

Force on - q, $\overrightarrow{\text{F}}=-q\overrightarrow{\text{E}}$

Magnitude of torque

$\tau=q\text{E}\times2a \text{ sin}\theta$

$\tau=2q\text{a}\times{\text{E}} \text{ sin}\theta$

$\overrightarrow\tau=\overrightarrow{\text{P}}\times\overrightarrow{\text{E}}$

2. If the electric field is non uniform, the dipole experiences a translatory force as well as a torque.

Net Electric Field at point $\text{P} =\int^{2\pi\text{a}}_{\circ}\text{dE}\cos\theta$

dE = Electric field due to a small element having charge dq 

$ = \frac{1}{4\pi\varepsilon_\circ}\frac{\text{dq}}{\text{r}^{2}}$

Let $\lambda$ = Linear charge density

$ = \frac{\text{dq}}{\text{dl}}$

$\text{dq} = \lambda\text{dl}$

Hence $\text{E} = \int^{2\pi\text{a}}_{\circ}\frac{1}{4\pi\varepsilon_\circ}.\frac{\lambda\text{dl}}{\text{r}^{2}}\times\frac{x}{r} , $

where $\cos\theta =\frac{\text{x}}{\text{r}}$

$ = \frac{\lambda\text{x}}{4\pi\varepsilon_\circ\text{r}^{3}}(2\pi\text{a})$

$ =\frac{1}{4\pi\varepsilon_\circ}\frac{\text{Qx}}{(\text{x}^{2} + \text{a}^{2})^{\frac{3}{2}}}$

 where total charge $\text{Q} = \lambda\times2\pi\text{a}$

At large distance i.e. x >> a

$\text{E}\simeq\frac{1}{4\pi\varepsilon_\circ}.\frac{\text{Q}}{\text{x}^{2}}$

This is the Electric field due to a point charge at distance x.

Electric field between the plates of capacitor E = $\frac{\sigma}{\varepsilon_{0}} =\frac{\text{Q}}{\text{A}\varepsilon_{0}}$

$\therefore$ potential difference

$\text{V} = \text{Ed} = \frac{\text{Qd}}{\text{A}\varepsilon_{0}}$

Capacitance

$\text{C} = \frac{\text{Q}}{\text{V}} = \frac{\varepsilon_{0}\text{A}}{\text{d}}$

2. When the two charged spherical conductors are connected by a conducting wire , they acquire the same potential

i.e $\frac{\text{Kq}_{1}}{\text{R}_{1}} = \frac{\text{Kq}_{2}}{\text{R}_{2}}\Rightarrow\frac{\text{q}_{1}}{\text{q}_{2}} = \frac{\text{R}_{1}}{\text{R}_{2}}$

Hence, ratio of surface charge densities

$\frac{\sigma_{1}}{\sigma_{2}} = \frac{\text{q}_{1}/4\pi\text{R}_{1}^{2}}{\text{q}_{2}/4\pi\text{R}_{2}^{2}}$

$ = \frac{\text{q}_{1}\text{R}_{2}^{2}}{\text{q}_{2}\text{R}_{1}^{2}}$

$ = \frac{\text{R}_{1}}{\text{R}_{2}}\times\frac{\text{R}_{2}^{2}}{\text{R}_{1}^{2}} = \frac{\text{R}_{2}}{\text{R}_{1}}.$

1. Initial voltage, V₁ = V volts and charge stored, Q₁ = 360 µC.

   Q₁ = CV₁ …(1)

   Changed potential, V₂ = V − 120

   Q₂ = 120 µC

   Q₂ = CV₂ ...(2)

   By dividing (2) from (1), we get  $\frac{\text{Q}_{1}}{\text{Q}_{2}}=\frac{\text{C}{V}_{1}}{\text{C}{V}_{2}} \Rightarrow \frac{360}{120}=\frac{\text{V}}{\text{V}-{120}}$

$\therefore {\text{C}}=\frac{\text{Q}_{1}}{\text{V}_{1}}=\frac{{360}\times{10}{^-}{^6}}{180}={2}\times{10}{^-}{^6} \text{F}={2}\mu{\text{F}}$

2.  If the voltage applied had increased by 120 V, then $\text{V}_{3}={180}+{120}={300}\text{V}.$

Hence, charge stored in the capacitor, $\text{Q}_{3}=\text{C}\text{V}_{3}={2}\times{10}{^-}{^6}\times{300}={600}\mu\text{C}.$

Given,  $\vec{E}={50}{x}\vec{i} \text{and}\bigtriangleup{s}={25}\text{C}\text{m}{^2}={25}\times{10}{^-}{^4}\text{m}{^2}$

As the electric field is only along the x-axis, so, flux will pass only through the cross-section of cylinder.

Magnitude of Electric Field at cross - section A, $\text{E}_\text{A}={50}\times{1}={50}\text{N}\text{C}{^-}{^1}$

Magnitude of Electric Field at cross - Section B, $\text{E}_\text{B}={50}\times{2}={100}\text{N}\text{C}{^-}{^1}$

The Corresponding Electric Fluxes are:

$\phi_\text{A}=\vec{\text{E}}.\bigtriangleup\vec{s}={50}\times{25}\times{10}{^-}{^4}\times\cos{180}{^0}=-{0.125}\text{N}\text{m}{^2}\text{C}{^-}{^1}$

$​​\phi_\text{B}=\vec{\text{E}}.\bigtriangleup\vec{s}={100}\times{25}\times{10}{^-}{^4}\times\cos{0}{^0}=-{0.25}\text{N}\text{m}{^2}\text{C}{^-}{^1}$

So, the net flux through the cylinder, $\phi=\phi_\text{A}+\phi_\text{B}= -{0.125} + {0.25}={0.125}\text{N}\text{m}{^2}\text{C}{^-}{^1}$

2.  Using Gauss’s law:

 $\oint\vec{\text{E}}{d}\vec{\text{s}}=\frac{\text{q}}{\in_{0}}\Rightarrow{0.125}=\frac{\text{q}}{{8.85}\times{10}{^-}{^1}{^2}}$

$\Rightarrow\text{q}={8.85}\times{0.125}\times{10}{^-}{^1}{^2}={1.1}\times{10}{^-}{^1}{^2}\text{C}$

From Gauss's theorem, ${\phi = \oint\overrightarrow{\text{E}}}.\text{d}\overrightarrow{\text{S}} = \frac{\text{q}_{m}}{\varepsilon_{0}}$

Flux $\phi$through S'.

$\phi = \oint\limits_{s'}\overrightarrow{\text{E}}.\text{d}\overrightarrow{\text{S}} = \oint\limits_{s'}\text{EdS} = \text{E}.4\pi\text{r}^{2}$

$\Rightarrow\text{E}.4\pi\text{r}^{2} = \frac{\text{q}_{m}}{\varepsilon_{0}}\Rightarrow\text{E} = \frac{1}{4\pi\varepsilon_{0}}\frac{\text{q}_{m}}{\text{r}^{2}}$

$\Rightarrow10\text{J}=(200-100)\text{v}\times\text{q}_0$

$\Rightarrow100\text{q}_0=10\text{v}$

$\Rightarrow\text{q}_0=\frac{10}{100}$

$=0.1\text{C}$

$\text{U}=\frac{1}{2}\in_0\text{E}^2$

$\text{F}=\frac{1}{4\pi\in_0}\frac{\text{q}_1\text{q}_2}{\text{r}^2}$

$\Rightarrow490=9\times10^9\times\frac{\text{q}^2}{1^2}$

$\Rightarrow\text{q}^2=54.4\times10^{-9}$

$\Rightarrow\text{q}=\sqrt{54.4\times10^{-9}}$

$=23.323\times10^{-5}\text{C}$

$\text{q}=2.3\times10^{-4}\text{C}$