Physics 3 Marks Question

1. Electric field inside a spherical conductor is zero. This is because if there is field inside the conductor, then charges will move to neutralize it.
2. Electric field E just outside the conductor is given by the relation,

$\text{E}=\frac{\text{q}}{4\pi\in_0\text{r}^2}$

Where,

∈₀ = Permittivity of free space

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

$\therefore\text{E}=\frac{1.6\times10^{-7}\times9\times10^{-9}}{(0.12)^2}$

= 10⁵ NC^-1

Therefore, the electric field just outside the sphere is 10⁵ NC^-1.

3. Electric field at a point 18 m from the centre of the sphere = E1

Distance of the point from the centre, d = 18 cm = 0.18 m

$\text{E}_1=\frac{\text{q}}{4\pi\in_0\text{d}^2}$

$=\frac{9\times10^9\times1.6\times10^{-7}}{18\times10^{-2}}$

= 4.4 × 10⁴ N/C

Therefore, the electric field at a point 18 cm from the centre of the sphere is 4.4 × 10⁴ N/C.

1. Potential at infinity is zero.

Potential energy of the system, p-e = Potential energy at infinity - Potential energy at distance, d

$=0-\frac{\text{q}_1\text{q}_2}{4\pi\in_0\text{d}}$

Where,

$\in_0$ is the permittivity of free space

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

$\therefore\text{Potential energy}=0-\frac{9\times10^9\times\big(1.6\times10^{-19}\big)^2}{0.53\times10^{-10}}=-43.7\times10^{-19}\text{J}$

Since 1.6x 10^-19 J = 1 eV,

$\therefore\text{Potential energy}=-43.7\times10^{-19}=\frac{-43.7\times10^{-19}}{1.6\times10^{-19}}=-27.2\ \text{eV}$

Therefore, the potential energy of the system is -27.2 ev.

2. Kinetic energy is half of the rnagnitude of potential energy.

Kinetic energy $=\frac{1}{2}\times(-27.2)=13.6\ \text{eV}$

Total energy = 13.6 - 27.2 = 13.6 eV

Therefore, the rninimum work required to free the electron is 13.6 eV.

3. When zero of potential energy is taken, d₁ = 1.06 A

$\therefore$ Potential energy of the system = Potential energy at d₁ - Potential energy at d

$=\frac{\text{q}_1\text{q}_2}{4\pi\in_0\text{d}_1}-27.2\ \text{eV}$

$=\frac{9\times10^9\times(1.6\times10^{-19})^2}{1.06\times10^{-10}}-27.2\ \text{eV}$

= 21.73 × 10^-19 J - 27.2 eV

= 13.58 eV - 27.2 eV

= -13.6 eV

$\frac{\text{E}_{\text{A}}}{\text{E}_{\text{B}}}=\frac{\text{Q}_{\text{A}}}{{4}\pi\in_{0} \text{a}^2}\times\frac{\text{b}^2{4}\pi\in_{0}}{\text{Q}_{\text{B}}}$

1. Initial capacitance, C = 1.771 × 10^-11

New capacitance, C^r = kC = 6 × 1.771 × 10^-11 = 106 pF

Supply voltage, V = 100 V

New charge q¹ = C' = 6 × 1.7717 × 10^-9 = 1.06 × 10^-8 Cs

Potential across the plates remains 100 V.

2. Dielectric constant, k = 6

Initial capacitance C' = kC = 6 × 1.771 × 10^-11 = 106 pF

If supply voltage is removed, then there will be no effect on the amount of charge in the plates.

Charge = 1.771 ×10^-9 C

Potential across the plates is given by,

$\therefore\text{V}'=\frac{\text{q}}{\text{C}'}$

$=\frac{1.771\times10^{-9}}{106\times10^{-12}}$

= 16.7 V

$\in_0$ = Permittivity of free space = 8.85 × 10^-12 N^-1 m^-2 C²

1. Charge placed at the centre of a shell is +q. Hence, a charge of magnitude -q will be induced to the inner surface of the shell. Therefore, total charge on the inner surface of the shell is -q.

Surface charge density at the inner surface of the shell is given by the relation,

$\sigma_1=\frac{\text{Total charge}}{\text{Inner surface area}}=\frac{-\text{q}}{4\pi\text{r}^2_1} \dots\dots(1)$

A charge of +q is induced on the outer surface of the shell. A charqe of magnitude Q is placed on the outer surface of the shell. Therefore, total charge on the outer surface of the shell is Q + q. Surface charge density at the outer surface of the shell,

$\sigma_2=\frac{\text{Total charge}}{\text{Outer surface area}}=\frac{-\text{Q+q}}{4\pi\text{r}^2_2} \dots\dots(2)$

2. Yes

The electric field intensity inside a cavity is zero, even if the shell is not spherical and has any irregular shape. Take a closed loop such that a part of it is inside the cavity along a field line while the rest is inside the conductor. Net work done by the field in carrying a test charge over a closed loop is zero because the field inside the conductor is zero. Hence, electric field is zero, whatever is the shape.

Distance between the two plates, d = 0.5cm = 0.5 × 10^-2m Capacitance of a parallel plate capacitor is given by the relation,

Where,

$\in_{0}= \text{Permittivity of free space} = {8.85}\times{10}^{-12}\text{C}^{2}\text{N}^{-1}\text{m}^{-2}$

$\therefore\ \text{A} = \frac{2\times0.5\times10^{-2}}{8.85\times10^{-12}}$

$={1130}\text{km}^{2}$

Hence, the area of the plates is too large. To avoid this situation, the capacitance is taken in the range of μF.

2. $\text{V} \propto\frac{1}{\text{r}}/$

3. No If the field lines are tangential, work will be done in moving a charge on the surface which goes against the definition of equipotential surface.

= (q/C) x dq

 $\therefore$Total energy stored in the capacitor

$\text{U} = \int\text{dw} = \int_{0}^{Q}\frac{\text{q}}{\text{c}}\text{dq} = \frac{1}{2}\frac{\text{Q}^{2}}{\text{c}}$

When battery is disconnected:

1. Energy stored will be decreased or energy stored = 1/K times the initial energy.
2. Electric field would decrease or E’ = E/K.

Alternate Answer

When battery is connected:

1. Energy stored will increase or become K times the initial energy.
2. Electric field will not change.

1. $\frac{1}{\text{C}} = \frac{1}{\text{C}_{1}} + \frac{1}{\text{C}_{2}} + \frac{1}{\text{C}_{3}}$

$\frac{1}{\text{C}} = \frac{1}{12} + \frac{1}{12} +\frac{1}{12} = \frac{1}{4}$

$\text{C} = 4\mu\text{F}\text{ or simply C}_\text{s} = \frac{\text{C}}{3} = \frac{12}{3}\mu\text{F} = 4 \mu\text{F}$

Equivalent capacitance

$\text{C}_{eq} = \text{C} + \text{C}_{4} = 4 +12 = 16 \mu\text{F}$

2. Calculation of charge on each capacitor:

Charge on capacitor C₄

$\text{Q}_{4} = \text{C}_{4}\text{V} = 12 \times500\mu\text{C} = 6000\mu\text{C} = 6 \times10^{-3}\text{C}$

Charge on capacitors C₁, C₂ and C₃

$\text{Q}_{123} =4 \mu\text{F}\times500\text{V} = 2 \times10^{-3}\text{C}. $

1. When an uncharged, grounded conducting plate is placed near a charged conducting plate a charge, of the opposite sign, gets induced on the
2. This reduces the potential of first plate, without any change in the charge present in it.

$\text{E} = \frac{\sigma}{\varepsilon_\circ}$

$\text{V}= \frac{\sigma}{\varepsilon_\circ}\text{d}$

$\text{C} =\frac{\varepsilon_\circ}{\text{d}}\text{A}$.

1. $\text{E}=\frac{1}{2}CV^2=\frac{6}{2}\times10^{-6}V^{2}=3\times10^{-6}V^2$

$\therefore V^2=\frac{E}{3\times10^{-6}}$

Energy stored in $12\mu f$ capacitor $=\frac{1}{2}CV^2$

$=\frac{1}{2}\times12\times10^{-6}\times\frac{E}{3\times10^{-6}}$

= 2E

2. Charge on $6\mu f$ capacitor, Q₁ $=\sqrt{2EC}\big[\because E= \frac{1}{2}\frac{Q^2}{C}\big]$

$=2\sqrt{3E}\times10^{-3}C$

Charge on $12\mu f$ capacitor, Q₂ $=\sqrt{2CE}$

$=\sqrt{2\times12\times10^{-6}\times2E}$

$=4\sqrt{3E}10^{-3}C$

Charge on $3\mu f$ apacitor, Q = Q₁+ Q₂

$=6\sqrt{3E}10^{-3}$

Energy stored in $3\mu f$ capacitor $=\frac{1}{2}\frac{Q^2}{C}=\frac{1}{2}\frac{36\times3E\times10^{-6}}{3\times10^{-6}}$

= 18E

Alternate Answer

2. capacitance of parallel combination = $18\mu f$

Charge on parallel combination, Q = CV

$=18\times10^{-6}V$

Charge on $3\mu f=\text{Q}=3\times10^{-6}V_1$

$=18\times10^{-6}V=3\times10^{-6}V_1$

$=V_1=6V$

$\therefore$ Energy stored in $3\mu f$ capacitor $=\frac{1}{2}CV_1^2$

$=\frac{1}{2}\times3\times10^{-6}\times\frac{E\times36}{3\times10^{-6}}$

= 18 E

3. Total energy drawn = E + 2E + 18E = 21E

In series combination: $\frac{1}{C_S}=\Bigg(\frac{1}{12}+\frac{1}{12}\Bigg)(pF)^{-1}$

$\therefore C_s=6\times10^{-12}pF$

$U_S=\frac{1}{2}CV^2$

$U_S=\frac{1}{2}\times6\times 10^{-12}\times50\times50\text{ }\text{J}$

$\therefore U_s=75\times10^{-10}\text{ }\text{J}$

$q_s=C_ sV$

$6\times50$

$300\times10^{-12} C=3\times{10}^{-10}C$

In parallel combination: ${C}_{p}=(12+12)pF$

$\therefore{C}_{p}=24\times10^{-12}F$

$U_s=\frac{1}{2}\times24\times10^{-12}\times2500\text{ }\text{J}$

$=3\times10^{-8}\text{ }\text{J}$

$q_p=C{p} V$

$q_p=24\times10{}^{-12}\times50\text{ }\text{C}$

$q_p=1.2\times10^{-9}\text{ }\text{C}$

1. Let $\text{C}_{x} = \text{C}$

$\text{C}_{y} = 4\text{C } \text{as it has a dielectric medium of } \varepsilon_{r} = 4 .$

For series combination of two capacitors

$\frac{1}{\text{C}} =\frac{1}{\text{C}_{x}} + \frac{1}{\text{C}_{y}}$

$\Rightarrow\frac{1}{4\mu\text{F}} = \frac{1}{\text{C}} + \frac{1}{4\text{C}}$

$\frac{1}{4\mu\text{F}} = \frac{5}{4\text{C}}$

$\Rightarrow\text{C} = 5\mu\text{F}$

Hence $\text{C}_{X} = 5\mu\text{F}$

$\text{C}_{y} = 20 \mu\text{F}$

2. Total charge Q = CV

$ = 4 \mu\text{F}\times15 \text{V} = 60 \mu\text{C}$

$\text{V}_{x} = \frac{\text{Q}}{\text{C}_{x}} = \frac{60\mu\text{C}}{5\mu\text{F}} = 12 \text{V}$

$\text{V}_{y} = \frac{\text{Q}}{\text{C}_{y}} = \frac{60\mu\text{C}}{20\mu\text{F}} = 3 \text{V}$

3. $\frac{{\text{E}}_{\text{x}}}{\text{E}_{\text{y}}} = \frac{^\frac{\text{Q}^{2}}{2{C}_{\text{X}}}}{\frac{\text{Q}^{2}}{2\text{C}_{Y}}} = \frac{\text{C}_{\text{y}}}{\text{C}_{\text{x}}} = \frac{20}{5} = 4:1$

$\text{E} = \frac{1}{2}\text{CV}^{2}$

$0.045 =\frac{1}{2}\frac{\text{c}_{1}\text{c}_{2}}{\text{c}_{1}\text{c}_{2}}(100)^{2}$

$ = > \frac{\text{c}_{1}\text{c}_{2}}{\text{c}_{1} + \text{c}_{2}} = 0.09\times10^{-4}$.....(i)

$0.25 = \frac{1}{2}(\text{c}_{1} +\text{c}_{2})(100)^{2}$

$= > \text{c}_{1} + \text{c}_{2} =0.5\times10^{-4}$.....(ii)

$\text{E} = \frac{1}{2}\text{CV}^{2}$

1. Capacltance $\text{C} = \frac{\text{K}\varepsilon_{0}\text{A}}{\text{d}},$ Hence capacitance Increases K times.
2. Potential difference $\text{V} = \frac{\text{V}_{o}}{\text{K}},$Hence potential difference decreases by a factor K.
3. Energy stored E $ =\frac{1}{2}\text{CV}^{2},$ As capacitance becomes K times & potential difference becomes 1/K times therefore energy stored becomes 1/K times.

Alternate Answer

Energy stored = Q²/2C. As capacitance increases by a factor K, the energy stored will decrease by the same factor.

$ = \text{q}_{2}\text{V}(\text{r}_{2}) + \frac{\text{q}_{1}\text{q}_{2}}{4\pi\varepsilon_{0}\text{r}_{12}}$

$ = \text{q}_{1}\text{V}(\text{r}_{1}) + \text{q}_{2}\text{V}(\text{r}_{2}) + \frac{\text{q}_{1}\text{q}_{2}}{4\pi\varepsilon_{0}\text{r}_{12}}.$

1. Charge on capacitors C₁ and C_2.

$\text{Q}_{1} = 36\mu\text{C} $ ; 

charge on capacitor C₃

_{$\text{Q}_{3} = 72\mu\text{C}$}

2. Equivalent capacitance.

$\text{C} = \frac{\text{C}_{1}{\text{C}_{2}}}{\text{C}_{1}+\text{C}_{2}} + \text{C}_{3}$

$\frac{6\times6}{6+6}+6 = 9\mu\text{F}$

3. Energy stored in Network.

$\text{W} =\frac{1}{2}\text{CV}^{2}$

$ =\frac{1}{2}\times9\times10^{-6}\times\big(12\big)^{2}\text{J}$

$ =648\times10^{-6}\text{J} = 648\mu\text{J}$

$ = 6.48\times10^{-4}\text{J}$

Potential at P due to +q

$\text{V}_{+q} = \frac{1}{4\pi\varepsilon_{\circ}}$ $\frac{q}{\times-a}$

Potential at P due to -q

$\text{V}_{-q} = \frac{1}{4\pi\varepsilon_{\circ}}$ $\frac{-q}{\times+a}$

$\therefore$ Potential at P due to the dipole

$\text{V} = \frac{1}{4\pi\varepsilon_{\circ}}$ $\frac{q\big(2\text{a}\big)}{(\text{X}^{2}-\text{a}^{2})}$

for a <<x , we have

$\therefore \text{v}\cong\frac{1}{4\pi\varepsilon_\circ}$ $\frac{q\big(2\text{a}\big)}{\text{X}^{2}}$

For a single charge V  $\alpha\frac{1}{\text{X}^{2}}$ 

1. No change.

As the battery is disconnected.

2. Decreases OR becomes $\frac{1}{\text{k}}$ times Due to polarisation of the dielectric.
3. Increases OR becomes k times.

As the electric field, and therefore, the p.d., between the plates decreases.

Force on q due to -4q.

$\text{F}_1=\frac{\text{K(4q)(q)}}{\text{l}^2}$ (along A - B)

$\text{F}_1=\frac{\text{4Kq}^2}{\text{l}^2}$ (along A - B)

Force on q due to 2q.

$\text{F}_2=\frac{\text{K(2q)(q)}}{\text{l}^2}$ (along C - A)

$\text{F}_2=\frac{2\text{Kq}^2}{\text{l}^2}$ (along C - A)

Angle between F₁ of F₂ $(\theta=120^\circ):$

$\therefore$ Resultat Force:

$\text{F}=\sqrt{\text{F}^2_1+\text{F}^2_2+2\text{F}_1\text{F}_2\cos120^\circ}$

$\text{F}=\sqrt{(2\text{F}_2)^2+(\text{F}_2)^2+2(2\text{F}_2)\text{F}_2\Big(-\frac{1}{2}\Big)}$ $(\because\ \text{F}_1=2\text{F}_2)$

$=\sqrt{5\text{F}^2_2-2\text{F}^2_2}=\sqrt{3}\text{F}_2$

$\therefore\ \text{F}=\sqrt{3}\times\frac{2\text{Kq}^2}{\text{l}^2}$

Direction $\tan\alpha=\frac{\text{F}_2\sin\theta}{\text{F}_1+\text{F}_2\cos\theta}$

$=\frac{\text{F}_2\times\frac{\sqrt{3}}{2}}{2\text{F}_2+\text{F}_2\big(-\frac{1}{2}\big)}$

So, resultant forces is making angle.

$\alpha=\tan^{-1}\Big(\frac{1}{\sqrt{3}}\Big)$ with the line AB.

2. Workdone to gather the charges from infinity = potential energy of system (U)

$\text{U}=\frac{\text{Kq(2q)}}{\text{l}}+\frac{\text{Kq(-4q)}}{\text{l}}+\frac{\text{K(2q)(-4q)}}{\text{l}}$

$=\frac{\text{Kq}^2}{\text{l}}(2-4-8)=-10\frac{\text{Kq}^2}{\text{l}}$

$\therefore$ Workdone to separate the charges to infinity $=\frac{10\text{Kq}^2}{\text{l}}$

Force on Q due to Q:

$\text{F}_1=\frac{\text{KQ}^2}{(\sqrt{2}\text{a})^2}=\frac{\text{KQ}^2}{2\text{a}^2}$ (direction is shown in figure)

Force on Q due to single 'q':

$\text{F}=\frac{\text{KQq}}{\text{a}^2}$ (direction is same as F₁)

Resultant Force on Q due to both 'q':

$\text{F}_2=\sqrt{2}\text{F}=\frac{\sqrt{2}\text{KQq}}{\text{a}^2}$

$\therefore$ Resulatant electric force on Q:

$\text{F}=\text{F}_1+\text{F}_2$

$\text{F}=\frac{\text{KQ}^2}{2\text{a}^2}+\frac{\sqrt{2}\text{KQq}}{\text{a}^2}$

$\text{F}=\frac{\text{KQ}}{\text{a}^2}\Big(\frac{\text{Q}}{2}+\sqrt{2}\text{q}\Big)$

$\text{F}=\frac{\text{KQ}}{\text{a}^2}\Big(\frac{\text{Q}+2\sqrt{2}\text{q}}{2}\Big)$

Directed outwards along the line joing charges Q and Q. (Show in figure)

2. Potential Energy of system:

$\text{U}=\frac{\text{KQq}}{\text{a}}+\frac{\text{KQq}}{\text{a}}+\frac{\text{KQ}^2}{\sqrt{2}\text{a}}+\frac{\text{KQq}}{\text{a}}+\frac{\text{Kq}^2}{\sqrt{2}\text{a}}+\frac{\text{KQq}}{\text{a}}$

$=\frac{4\text{KQq}}{\text{a}}+\frac{\text{KQ}^2}{\sqrt{2}\text{a}}+\frac{\text{Kq}^2}{\sqrt{2}\text{a}}$

$\text{P.E.}=\frac{\text{K}}{\text{a}}\Big[4\text{Qq}+\frac{\text{Q}^2}{\sqrt{2}}+\frac{\text{q}^2}{\sqrt{2}}\Big]$​​​​​​​

1. For constant electric field vector E

For increasing electric field

Difference: For constant electric field, the equipotential surfaces are equidistant for the same potential difference between these surfaces; while for increasing electric field, the separation between these surfaces decreases, in the direction of the increasing field, for the same potential difference between them.

2. Suppose P is a point on the axial position of the dipole. Length of dipole = 2a

Suppose point P is at the distance 'r' from the center of the dipole.

We know that potential at a point is given as, $\text{V}=\frac{1}{4\pi\in_0}.\frac{\text{Q}}{\text{r}}$

So, the potential at P due to q is, $\text{V}_\text{q}=\frac{1}{4\pi\in_0}.\frac{\text{q}}{\text{a}+\text{r}}$

Potential at P due to -q is, $\text{V}_\text{-q}=\frac{1}{4\pi\in_0}.\frac{\text{-q}}{\text{a}-\text{r}}$

The total potential at P is,

$\text{V}=\text{V}_\text{q}+\text{V}_{-\text{q}}$

$=\frac{1}{4\pi\in_0}.\frac{\text{q}}{(\text{a}+\text{r})}+\frac{1}{4\pi\in_0}.\frac{-\text{q}}{(\text{r-a})}$

$=\frac{\text{q}}{4\pi\in_0}\Big[\frac{1}{(\text{a+r})}+\frac{1}{(\text{a-r})}\Big]$

$\text{V}=\frac{\text{q}}{4\pi\in_0}.\frac{2\text{a}}{(\text{a}^2-\text{r}^2)}$

$\text{V}=\frac{\text{q}_1\text{q}_2}{4\pi\in_0\text{d}_1}+\frac{\text{q}_2\text{q}_3}{4\pi\in_0\text{d}_1}$

$\text{C}'=\frac{\text{C}_1\text{C}_2}{\text{C}_2+\text{C}_2}=\frac{2\times2}{2+2}=1\mu\text{F}$

$\text{C}''=1+1=2\mu\text{F}$

$\text{C}'''=\frac{\text{C}_4\text{C}}{\text{C}_4+\text{C}''}=\frac{2\times2}{2+2}=1\mu\text{F}$