Physics 3 Marks Question

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

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

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

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

$=1.03\times10^{-5}$

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.

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

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

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

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

1. $\text{A}=(0,0);\ \text{B}=(4,2)$

$\Rightarrow\text{V}_\text{B}-\text{V}_\text{A}=\text{E}\times\text{d}$

$=20\times\sqrt{16}$

$=80\text{V}$

2. $\text{A}=(4\text{m},2\text{m});\ \text{B}=(6\text{m},5\text{m})$

$\Rightarrow\text{V}_\text{B}-\text{V}_\text{A}=\text{E}\times\text{d}$

$=20\times\sqrt{(6-4)^2}$

$=20\times2$

$=40\text{V}$

3. $\text{A}=(0,0);\ \text{B}=(6\text{m},5\text{m})$

$\Rightarrow\text{V}_\text{B}-\text{V}_\text{A}=\text{E}\times\text{d}$

$=20\times\sqrt{(6-0)^2}$

$=20\times6$

$=120\text{V}.$

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

$=24.94\text{N}.$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$=0.6\times10^{-2}$

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

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

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

$=0.005\text{kg}$

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

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

$=14.4\text{N}$

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.

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

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

But $\text{qE}=\text{mg}$

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

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

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

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

$=6.6\times10^{3}$