Electric Charges MCQ Questions & Answers in Electrostatics and Magnetism | Physics

I
$$\eqalign{ & {F_e}\cos \theta = mg\sin \theta \cr & \Rightarrow \frac{{{F_e}}}{{mg}} = \tan \theta \cr} $$

$$ \Rightarrow mg = \frac{{{F_e}}}{{\tan \theta }}\,......\left( {\text{i}} \right)$$
II
$$\eqalign{ & {F_e}^\prime \cos {\theta _1} = mg\sin {\theta _1} \cr & \Rightarrow \frac{{{F_e}^\prime }}{{mg}} = \tan {\theta _1} \cr & \Rightarrow mg = \frac{{{F_e}^\prime }}{{\tan {\theta _1}}}\,......\left( {{\text{ii}}} \right) \cr} $$
From Eqs. (i) and (ii), we get
$$\frac{{{F_e}}}{{\tan \theta }} = \frac{{{F_e}^\prime }}{{\tan {\theta _1}}} \Rightarrow \frac{{{F_e}}}{{{F_e}^\prime }} = \frac{{\tan \theta }}{{\tan {\theta _1}}}$$
As, $$\tan \theta = \frac{{P\left( {{\text{perpendicular}}} \right)}}{{b\left( {{\text{base}}} \right)}} = \frac{r}{{2y}}$$
$$\tan {\theta _1} = \frac{P}{b} = \frac{r}{{2 \times y}} \times 2 = \frac{r}{y}$$

$$\eqalign{ & \frac{{{F_e}}}{{{F_e}^\prime }} = \frac{r}{{2y \times r}} \times y \cr & \Rightarrow {F_e}^\prime = 2{F_e} \cr & {\text{So,}}\,\,\frac{{k{q^2}}}{{{{r'}^2}}} = \frac{{2k{q^2}}}{{{r^2}}} \cr & \frac{1}{{r'}} = \frac{{\sqrt 2 }}{r} \cr & \therefore r' = \frac{r}{{\sqrt 2 }} \cr} $$

Here, $${q_1} = {q_2} = 3.2 \times {10^{ - 7}}C,r = 60\,cm = 0.6\,m$$
Electrostatic force, $$F = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{q_1}{q_2}}}{{{r^2}}}$$
$$ = \frac{{9 \times {{10}^9}{{\left( {3.2 \times {{10}^{ - 7}}} \right)}^2}}}{{{{\left( {0.6} \right)}^2}}} = 2.56 \times {10^{ - 3}}N$$

Electrostatic force is given by
$$F = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{q_1}{q_2}}}{{{r^2}}}$$
Here, charge and distance are same. So, force between two protons will be same.

$$\eqalign{ & {\text{We}}\,{\text{know,}}\,{V_0} = \frac{{Kq}}{R} = V\,surface \cr & {\text{Now,}}\,{V_i} = \frac{{Kq}}{{2{R^3}}}\left( {3{R^2} - {r^2}} \right)\,\,\,\left[ {{\text{For}}\,r < R} \right] \cr} $$
At the centre of sphare $$r = 0.$$  Here
$$\eqalign{ & V = \frac{3}{2}{V_0} \cr & {\text{Now,}}\,\,\frac{5}{4}\frac{{Kq}}{R} = \frac{{Kq}}{{2{R^3}}}\left( {3{R^2} - {r^2}} \right) \cr & {R_2} = \frac{R}{{\sqrt 2 }} \cr & \frac{3}{4}\frac{{Kq}}{R} = \frac{{Kq}}{{{R^3}}} \cr & \frac{1}{4}\frac{{Kq}}{R} = \frac{{Kq}}{{{R_4}}} \cr & {R_4} = 4R \cr & {\text{Also,}}\,{R_1} = 0\,{\text{and}}\,{R_2} < \left( {{R_4} - {R_3}} \right) \cr} $$

$$\eqalign{ & {F_{{\text{net}}}} = 2\left| {{F_{31}}} \right|\cos \alpha \cr & = 2 \times \frac{1}{{4\pi {\varepsilon _0}}} \times \frac{{2 \times 4 \times {{10}^{ - 12}}}}{{{{\left( {0.5} \right)}^2}}} \times \frac{4}{5} = 0.46\,N \cr} $$

Force on charge $${q_1}$$ due $${q_2}$$ is $${F_{12}} = k\frac{{{q_1}{q_2}}}{{{b^2}}}$$
Force on charge $${q_1}$$ due $${q_3}$$ is $${F_{13}} = k\frac{{{q_1}{q_3}}}{{{a^2}}}$$
The $$X$$-component of the force $$\left( {{F_x}} \right)$$ on $${q_1}$$ is $${F_{12}} + {F_{13}}\sin \theta $$
$$\eqalign{ & \therefore {F_x} = k\frac{{{q_1}{q_2}}}{{{b^2}}} + k\frac{{{q_1}{q_2}}}{{{a^2}}}\sin \theta \cr & \therefore {F_x} \propto \frac{{{q_2}}}{{{b^2}}} + \frac{{{q_3}}}{{{a^2}}}\sin \theta \cr} $$

$$\frac{1}{2}m{v^2} = \frac{{kQq}}{r} \Rightarrow \frac{1}{2}m{\left( {2v} \right)^2} = \frac{{kqQ}}{{r'}} \Rightarrow r' = \frac{r}{4}$$

Number of protons $$ = \frac{{1C}}{{1.66 \times {{10}^{ - 19}}C}} \approx 6 \times {10^{18}}$$

Here we have $$\frac{{Qq}}{a} + \frac{{{q^2}}}{a} + \frac{{Qq}}{{a\sqrt 2 }} = 0$$
$$\therefore Q = - \frac{{q\sqrt 2 }}{{\sqrt 2 + 1}} = - \frac{{2q}}{{2 + \sqrt 2 }}$$

In equilibrium, $${F_e} = T\sin \theta $$
$$\eqalign{ & mg = T\cos \theta \cr & \tan \theta = \frac{{{F_e}}}{{mg}} = \frac{{{q^2}}}{{4\pi { \in _0}{x^2} \times mg}} \cr & {\text{also}}\,\tan \theta \approx \sin = \frac{{\frac{x}{2}}}{l} \cr & {\text{Hence,}}\,\frac{x}{{2l}} = \frac{{{q^2}}}{{4\pi { \in _0}{x^2} \times mg}} \cr & \Rightarrow {x^3} = \frac{{2{q^2}l}}{{4\pi { \in _0}mg}} \cr & \therefore x = {\left( {\frac{{{q^2}l}}{{2\pi { \in _0}mg}}} \right)^{\frac{1}{3}}} \cr & \Rightarrow x \propto {l^{\frac{1}{3}}} \cr} $$