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

Two positive ions each carrying a charge $$q$$ are kept at a distance $$d,$$ then it is found that force of repulsion between them is $$F = \frac{{kqq}}{{{d^2}}} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{qq}}{{{d^2}}}$$
where, $$q = ne$$
$$\eqalign{ & \therefore F = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{n^2}{e^2}}}{{{d^2}}} \cr & \Rightarrow n = \sqrt {\frac{{4\pi {\varepsilon _0}F{d^2}}}{{{e^2}}}} \cr} $$

$$\eqalign{ & {\text{Here,}}\,{q_1} = 1 \times {10^{ - 7}}C,\,\,{\text{and}}\,{q_2} = 2 \times {10^{ - 7}}C, \cr & r = 20\,cm = 20 \times {10^{ - 2}}m \cr & F = \frac{{{q_1}{q_2}}}{{4\pi {\varepsilon _0}{r^2}}} = \frac{{9 \times {{10}^9} \times 1 \times {{10}^{ - 7}} \times 2 \times {{10}^{ - 7}}}}{{{{\left( {20 \times {{10}^{ - 2}}} \right)}^2}}} \cr & = 4.5 \times {10^{ - 3}}N \cr} $$

At any instant
$$\eqalign{ & T\cos \theta = mg\,......\left( {\text{i}} \right) \cr & T\sin \theta = {F_e}\,......\left( {{\text{ii}}} \right) \cr & \Rightarrow \frac{{\sin \theta }}{{\cos \theta }} = \frac{{{F_e}}}{{mg}} \Rightarrow {F_e} = mg\tan \theta \cr & \Rightarrow \frac{{k{q^2}}}{{{x^2}}} = mg\tan \theta \Rightarrow {q^2} \propto {x^2}\tan \theta \cr & \sin \theta = \frac{x}{{2l}} \cr} $$
For small $$\theta ,\sin \theta \approx \tan \theta \,\,\,\,\therefore {q^2} \propto {x^3}$$

$$\eqalign{ & \Rightarrow q\frac{{dq}}{{dt}}\, \propto \,{x^2}\frac{{dx}}{{dt}} \cr & \therefore \frac{{dq}}{{dt}} = {\text{const}}{\text{.}} \cr & \therefore q\, \propto \,{x^2}.v \Rightarrow {x^{\frac{3}{2}}}\alpha {x^2}.v\left[ {\because {q^2}\, \propto \,{x^3}} \right] \cr & \Rightarrow v\, \propto \,{x^{ - \frac{1}{2}}} \cr} $$

The bowl exerts a normal force $$N$$ on each bead, directed along the radial line or at $${60^ \circ }$$ above the horizontal. Consider the free-body diagram of the bead on the left with the electric force $${F_e}$$ applied:

$$\eqalign{ & \Sigma {F_y} = N\sin {60^ \circ } - mg = 0 \cr & \Rightarrow N = \frac{{mg}}{{\sin {{60}^ \circ }}} \cr & \Sigma {F_x} = - {F_e} + N\cos {60^ \circ } = 0 \cr & \Rightarrow \frac{{{k_e}{q^2}}}{{{R^2}}} = N\cos {60^ \circ } = \frac{{mg}}{{\tan {{60}^ \circ }}} = \frac{{mg}}{{\sqrt 3 }} \cr & {k_e} = \frac{1}{{4\pi {\varepsilon _0}}} \approx 9.0 \times {10^9}N{m^2}/{C^2} \cr & \therefore q = R{\left( {\frac{{mg}}{{{k_e}\sqrt 3 }}} \right)^{\frac{1}{2}}} \cr} $$

$$\eqalign{ & T\sin \theta = \frac{\sigma }{{{\varepsilon _0}K}}.q\,.......\left( {\text{i}} \right) \cr & T\cos \theta = mg\,.......\left( {{\text{ii}}} \right) \cr} $$
Dividing (i) by (ii),
$$\eqalign{ & \tan \theta = \frac{{\sigma q}}{{{\varepsilon _0}K.mg}} \cr & \therefore \sigma \propto \tan \theta \cr} $$

$$\eqalign{ & \frac{{{W_{PQ}}}}{q} = \left( {{V_Q} - {V_P}} \right) \cr & \Rightarrow {W_{PQ}} = q\left( {{V_Q} - {V_P}} \right) \cr & = \left( { - 100 \times 1.6 \times {{10}^{ - 19}}} \right)\left( { - 4 - 10} \right) \cr & = + 2.24 \times {10^{ - 16}}J \cr} $$

$$F \propto \frac{{{Q_A}{Q_C}}}{{{x^2}}}$$
$$x$$ is distance between the spheres. After first operation charge on $$B$$ is halved i.e $$\frac{Q}{2}$$ and charge on third sphere becomes $$\frac{Q}{2}.$$  Now it is touched to $$C,$$ charge then equally distributes them selves to make potential same,hence charge on $$C$$ becomes $$\left( {Q + \frac{Q}{2}} \right)\frac{1}{2} = \frac{{3Q}}{4}.$$
$$\eqalign{ & \therefore {F_{new}} \propto \frac{{Q{'_C}Q{'_B}}}{{{x^2}}} = \frac{{\left( {\frac{{3Q}}{4}} \right)\left( {\frac{Q}{2}} \right)}}{{{x^2}}} = \frac{3}{8}\frac{{{Q^2}}}{{{x^2}}} \cr & {\text{or}}\,{F_{new}} = \frac{3}{8}F \cr} $$

Each induce other oppositely near by.

Let $$n$$ be the number of electrons missing.
$$\eqalign{ & F = \frac{1}{{4\pi {\varepsilon _0}}}.\frac{{{q^2}}}{{{d^2}}} \Rightarrow q = \sqrt {4\pi {\varepsilon _0}{d^2}F} = ne \cr & \therefore n = \sqrt {\frac{{4\pi {\varepsilon _0}F{d^2}}}{{{e^2}}}} \cr} $$

For external points, a charged sphere behaves as if the whole of its charge is concentrated at its centre.
Force on $$A$$ due to $$B,$$
$$\eqalign{ & {F_{AB}} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{q^2}}}{{{{\left( {2R} \right)}^2}}} \cr & = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{q^2}}}{{4{R^2}}}\,{\text{along}}\,\overrightarrow {BA} \cr} $$
And force on $$A$$ due to $$C,$$
$${F_{AC}} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{q^2}}}{{{{\left( {2R} \right)}^2}}} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{q^2}}}{{4{R^2}}}\,{\text{along}}\,\overrightarrow {CA} $$
Now as angle between $$BA$$ and $$CA$$ is $${60^ \circ }$$ and
$$\left| {{F_{AB}}} \right| = \left| {{F_{AC}}} \right| = F$$
$$\eqalign{ & \therefore {F_A} = \sqrt {{F^2} + {F^2} + 2FF\cos {{60}^ \circ }} = \sqrt 3 F \cr & = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{\sqrt 3 }}{4}{\left( {\frac{q}{R}} \right)^2} \cr} $$