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

From Coulomb’s law $$F = \frac{{K{q_1}{q_2}}}{{{r^2}}}$$
i.e., $$F \propto \frac{1}{{{r^2}}}$$
which is correctly shown by graph (D).

They will not experience any force if
$$\eqalign{ & \left| {{{\vec F}_G}} \right| = \left| {{{\vec F}_e}} \right| \cr & \Rightarrow G\frac{{{m^2}}}{{{{\left( {16 \times {{10}^{ - 2}}} \right)}^2}}} = \frac{1}{{4\pi {\varepsilon _0}}}.\frac{{{q^2}}}{{{{\left( {16 \times {{10}^{ - 2}}} \right)}^2}}} \cr & \Rightarrow \frac{q}{m} = \sqrt {4\pi {\varepsilon _0}G} \cr} $$

$$\eqalign{ & {V_A} = {V_{self}} + {V_{due\,to\,\left( 2 \right)}} \cr & \Rightarrow {V_A} = \frac{1}{{4\pi {\varepsilon _0}}}\left[ {\frac{q}{R} - \frac{q}{{\sqrt {{R^2} + {d^2}} }}} \right] \cr & {V_B} = {V_{self}} + {V_{due\,to\,\left( 1 \right)}} \cr & \Rightarrow {V_B} = \frac{1}{{4\pi {\varepsilon _0}}}\left[ {\frac{{ - q}}{R} + \frac{q}{{\sqrt {{R^2} + {d^2}} }}} \right] \cr & \Delta V = {V_A} - {V_B} \cr & = \frac{1}{{4\pi {\varepsilon _0}}}\left[ {\frac{q}{R} + \frac{q}{R} - \frac{q}{{\sqrt {{R^2} + {d^2}} }} - \frac{q}{{\sqrt {{R^2} + {d^2}} }}} \right] \cr & = \frac{1}{{2\pi {\varepsilon _0}}}\left[ {\frac{q}{R} - \frac{q}{{\sqrt {{R^2} + {d^2}} }}} \right] \cr} $$

From figure, $$\tan \theta = \frac{{{F_e}}}{{mg}} \Rightarrow \frac{{\frac{r}{2}}}{y} = \frac{{\frac{{k{q^2}}}{{{r^2}}}}}{{mg}}\,\,\left[ {\because F = \frac{{k{q^2}}}{{{r^2}}}\,{\text{from coulomb's law}}} \right]$$
$$ \Rightarrow {r^3} \propto y \Rightarrow {{r'}^3} \propto \frac{y}{2} = \frac{{r'}}{r} = \frac{1}{{{2^{\frac{1}{3}}}}} \Rightarrow r' = \frac{r}{{\root 3 \of 2 }}$$

Electronic charge does not depend on acceleration due to gravity as it is a universal constant.
So, electronic charge on earth = electronic charge on moon
$$\therefore $$ Required ratio = 1.

$$\eqalign{ & {\text{Here, }}V\left( x \right) = \frac{{20}}{{{x^2} - 4}}{\text{ }}volt \cr & {\text{We know that }}E = - \frac{{dV}}{{dx}} = - \frac{d}{{dx}}\left( {\frac{{20}}{{{x^2} - 4}}} \right) \cr & {\text{or, }}E = + \frac{{40x}}{{{{\left( {{x^2} - 4} \right)}^2}}} \cr & {\text{At }}x = 4\mu m \cr & E = + \frac{{40 \times 4}}{{{{\left( {{4^2} - 4} \right)}^2}}} = + \frac{{160}}{{144}} = + \frac{{10}}{9}{\text{ }}volt{\text{ }}/\mu m \cr} $$

On touching the metal knob with a positively charged rod some electrons from the gold leaves get transferred to the rod making gold leaves positively charged and they get separated. When a negatively charged rod is touched with metal knob some negative charge flows to the gold leaves lessening the positive charge there and the separation between the leaves decreases.

$$\eqalign{ & {T_{AB}} = \frac{{kq \cdot 2q}}{{{d^2}}} + \frac{{kq \cdot 4q}}{{{{\left( {2d} \right)}^2}}} = \frac{{3k{q^2}}}{d} \cr & {T_{BC}} = \frac{{k \cdot 4q \cdot 2q}}{{{d^2}}} + \frac{{kq \cdot 4q}}{{{{\left( {2d} \right)}^2}}} = \frac{{9k{q^2}}}{{{d^2}}} \cr} $$

We know that $$\frac{{{W_{AB}}}}{q} = {V_B} - {V_A}$$
$$\therefore {V_B} - {V_A} = \frac{{2J}}{{20C}} = \frac{{0.1J}}{C} = 0.1V$$

$$\eqalign{ & U = \frac{1}{{4\pi { \in _0}}}\left[ {\frac{{q\left( { - q} \right)}}{d} + \frac{{qQ}}{{\left( {D + \frac{d}{2}} \right)}} + \frac{{\left( { - q} \right)Q}}{{\left( {D - \frac{d}{2}} \right)}}} \right] \cr & = \frac{1}{{4\pi {\varepsilon _0}}}\left[ {\frac{{ - {q^2}}}{d} + \frac{{qQ\left( {D - \frac{d}{2}} \right) - qQ\left( {D + \frac{d}{2}} \right)}}{{{D^2} - \frac{{{d^2}}}{4}}}} \right] \cr & = \frac{1}{{4\pi { \in _0}}}\left[ { - \frac{{{q^2}}}{d} - \frac{{qQd}}{{{D^2}}}} \right]\because \frac{{{d^2}}}{4} < < D \cr} $$