Physics 3 Marks Question

$\text{q}_1=\text{q}_2=\text{q}=1.0\text{C}$

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

$\in_0$ = parmittivity of free space.

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

$\text{F}=\text{qE}$

$\text{F}=\frac{\text{q}\times\sigma}{2\in_0}$

$\text{F}=\frac{\big(2.0\times10^{-6}\big)\times\big(4.0\times10^{-6}\big)}{2\times\big(8.85\times10^{-12}\big)}$

$\text{F}=0.45\text{N}$

$\text{T}=2\pi\sqrt{\frac{\ell}{\text{g}}}$ $\big(\ell=$ length, q = acceleration$\big)$

$\vec{\text{E}}=\lim\limits_{\text{q}_0\rightarrow0}\frac{\vec{\text{F}}}{\text{q}_0}$

Electric Field Strength at mid-point of dipole: The electric field strength at mid-point C due to charge +q is -q along the same direction.

$=\frac{1}{4\pi\in_0}\frac{\text{q}}{\text{a}^2}+\frac{1}{4\pi\in_0}\frac{\text{q}}{\text{a}^2}=\frac{1}{4\pi\in_0}\frac{2\text{q}}{\text{a}^2}$

Three charges are held at three corners of a equilateral trangle.

Let the charges be A, B and C.

It is of length 5cm or 0.05m

Force exerted by B on A = F₁

Force exerted by C on A = F₂

So, force exerted on A = resultant F₁ = F₂

$\Rightarrow\text{F}=\frac{\text{kq}_2}{\text{r}^2}$

$=\frac{9\times10^9\times2\times2\times2\times10^{-12}}{5\times5\times10^{-4}}$

$=\frac{36}{25}\times10$

$=14.4$

Now, force on A = 2 × F cos 30^° since it is equilateral $\triangle.$

⇒ Force on $\text{A}=2\times1.44\times\sqrt{\frac{3}{2}}$

$=24.94\text{N}.$

$\text{m}=10,\ \text{mg}=10\times10^{-3}\text{g}\times10^{-3}\text{kg},$

$\text{q}=1.5\times10^{-6}\text{C}$

But $\text{qE}=\text{mg}$

$\Rightarrow(1.5\times10^{-6})\text{E}=10\times10^{-6}\times10$

$\Rightarrow\text{E}=\frac{10\times10^{-4}\times10}{1.5\times10^{-6}}$

$=\frac{100}{1.5}=66.6\text{N/C}$

$=\frac{100\times10^3}{1.5}=\frac{10^{5+1}}{15}$

$=6.6\times10^{3}$

Let -q & -q are placed at A & C

Where 2q on B

So length of A = d

So the dipole moment = (q × d) = P

So, Resultant dipole moment

$\text{P}=\Big[(\text{qd})^2+(\text{qd})^2+2\text{qd}\times\text{qd}\cos60^\circ\Big]^{\frac{1}{2}}$

$=\big[3\text{q}^2\text{d}^2\big]^{\frac{1}{2}}$

$=\sqrt{3}\text{qd}$

$=\sqrt{3}\text{p}$

1. At A, $\text{E}=\frac{\sigma}{2\in_0}=\frac{2\times10^{17}\text{Cm}^{-2}}{2\times8.854\times10^{-12}\text{C}^2\text{N}^{-1}\text{m}^{-2}}$

$\text{E}=1.1\times1028 \text{N/C}$

Directed away from the sheet.

2. Point Y, Because at 50cm, the charge sheet acts as a finite sheet and thus the magnitude remains same towards the middle region of the planar sheet.

Mass of the bob = 100g = 0.1kg

So Tension in the string = 0.1 × 9.8 = 0.98N.

For the Tension to be 0, the charge below should repel the first bob.

$\Rightarrow\text{F}=\frac{\text{kq}_1\text{q}_2}{\text{r}^2}$ $\big[\text{T}-\text{mg}+\text{F}=0\ \Rightarrow\text{T}=\text{mg}-\text{f},\ \text{T}=\text{mg}\big]$

$\Rightarrow0.98=\frac{9\times10^9\times2\times10^{-4}\times\text{q}^2}{(0.01)^2}$

$\Rightarrow\text{q}_2=\frac{0.98\times1\times10^{-2}}{9\times2\times10^5}$

$=0.054\times10^{-9}\text{N}$

$\text{q}_1=\text{q}_2=4\times10^{-5}$

$\text{s}=1\text{m},\ \text{m}=5\text{g}$

$=0.005\text{kg}$

$\text{F}=\text{K}\frac{\text{q}^2}{\text{r}^2}$

$=\frac{9\times10^9\times\big(4\times10^{-5}\big)^2}{1^2}$

$=14.4\text{N}$

Acceleration ‘a’ $=\frac{\text{F}}{\text{m}}$

$=\frac{14.4}{0.005}$

$=2880\text{m/s}^2$

Now, $\text{u}=0,$

$\text{s}=50\text{cm}=0.5\text{m}$

$\text{a}=2880\text{m/s}^2,\ \text{V}=?$

$\Rightarrow\text{V}=\sqrt{2880}$

$=53.66\text{m/s}\approx54\text{m/s}$ for each particle.

$\oint\overrightarrow{\text{E}}.\text{d}\overrightarrow{\text{S}}=\frac{1}{\in_0}\int\rho\text{dV}$

$\frac{\text{q}_1}{\text{q}_2}=\frac{6}{18}$

$\Rightarrow\frac{\text{q}_1}{\text{q}_2}=\frac{1}{3}$

$\text{m}=10\text{g}$

$\text{F}=\frac{\text{KQ}}{\text{r}}$

$=\frac{9\times10^9\times2\times10^{-4}}{10\times10^{-2}}$

$\text{F}=1.8\times10^{-7}$

$\text{F}=\text{m}\times\text{a}$

$\Rightarrow\text{a}=\frac{1.8\times10^{-7}}{10\times10^{-3}}$

$=1.8\times10^{-3}\text{m/s}^2$

$\text{V}^2-\text{u}^2=2\text{as}$

$\Rightarrow\text{V}^2=\text{u}^2+2\text{as}$

$\text{V}=\sqrt{0+2\times1.8\times10^{-3}\times10\times10^{-2}}$

$=\sqrt{3.6\times10^{-4}}$

$=0.6\times10^{-2}$

$=6\times10^{-3}\text{,m/s}.$

$\text{G}=50\mu\text{C}=50\times10^{-6}\text{C}$

We have, $\text{E}=\frac{2\text{KQ}}{\text{r}}$ for a charged cylinder

$\Rightarrow\text{E}=\frac{2\times9\times10^9\times50\times10^{-6}}{5\sqrt{3}}$

$=\frac{9\times10^{-5}}{5\sqrt{3}}$

$=1.03\times10^{-5}$

Charge q is in equilibrium since charges A and B exert equal and opposite forces on it.

For equilibrium of charge Q at B;

F_BC + F_AB = 0

$\Rightarrow\ \frac{1}{4\pi\in_0}\frac{\text{qQ}}{(\text{l}/2)^2}+\frac{1}{4\pi\in_0}\frac{\text{Q.Q}}{\text{l}^2}=0$

$\Rightarrow\ \frac{1}{4\pi\in_0}\frac{\text{Q}}{\text{l}^2}(4\text{q}+\text{Q})=0\Rightarrow\ \text{}q=-\frac{\text{Q}}{4}$

Electric flux through sphere S₁, $\Phi_1=\frac{2(\text{Q})}{\in_0}$

Electric flux through sphere S₂, $\Phi=\frac{(2\text{Q}+4\text{Q})}{\in_0}=\frac{6\text{Q}}{\in_0}$

$\text{Ratio}=\frac{\Phi_1}{\Phi}=\frac{\frac{2\text{Q}}{\in_0}}{\frac{6\text{Q}}{\in_0}}=\frac{1}{3}$

If a medium of dielectric constant $\text{K}(=\in_\text{r})$ is filled in the sphere S₁, electric flux through sphere,

$\Phi'_1=\frac{2\text{Q}}{\in_\text{r}\in_0}=\frac{2\text{Q}}{\text{K}\in_0}.$

Consider the Gaussian surface as shown in the figure.

Let the charge on the outer surface of the left-most plate be q. Thus, the charges on the plates are distributed as shown in the diagram.

The net field at point P due to all the induced charges must be zero, as it is lying inside the metal surface.

Let the surface area of the plates be A.

Electric field at point P due to the charges on plate X:

Due to charge (+Q - q) is $\frac{\text{Q}-\text{q}}{2\text{A}\in_0}$ in the right direction

Due to charge (+q) is $\frac{\text{q}}{2\text{A}\in_0}$ in the right direction

Electric field at point P due to charges on plate Y:

Due to charge (-q) is $\frac{\text{q}}{2\text{A}\in_0}$ in the right direction

Due to charge (+q) is $\frac{\text{q}}{2\text{A}\in_0}$ in the left direction

Electric field at point P due to charges on plate Z:

Due to charge (-q) is $\frac{\text{q}}{2\text{A}\in_0}$ in the right direction

Due to charge (-2Q + q) is $\frac{\text{2Q}-\text{q}}{2\text{A}\in_0}$ in the right direction

The net electric field at point P:

$\frac{\text{Q}-\text{q}}{2\text{a}\in_0}+\frac{\text{q}}{2\text{A}\in_0}-\frac{\text{q}}{2\text{A}\in_0}-\frac{\text{q}}{2\text{A}\in_0}+\frac{\text{q}}{2\text{A}\in_0}+\frac{\text{2Q}-\text{q}}{2\text{A}\in_0}=0$

$\frac{\text{Q}-\text{q}}{2\text{A}\in_0}+\frac{\text{2Q}-\text{q}}{2\text{A}\in_0}=0$

$\text{Q}-\text{q}+\text{2Q}-\text{q}=0$

$\text{3Q}-2\text{q}=0$

$\text{q}=\frac{\text{3Q}}{2}$

Thus, the charge on the outer plate of the right-most plate

$-2\text{Q}+\text{q}=-2\text{Q}+\frac{3\text{Q}}{2}=-\frac{\text{Q}}{2}$