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

Net charge on one $$H$$-atom $$q = - e + e + \Delta e = \Delta e$$
Net electrostatic repulsive force between two $$H$$-atoms $${F_r} = \frac{{K{q^2}}}{{{d^2}}} = \frac{{K{{\left( {\Delta e} \right)}^2}}}{{{d^2}}}$$
Similarly, net gravitational attractive force between two $$H$$-atoms $${F_G} = \frac{{Gm_r^2}}{{{d^2}}}$$
It is given that $${F_r} - {F_G} = 0$$
$$\eqalign{ & \Rightarrow \frac{{K{{\left( {\Delta e} \right)}^2}}}{{{d^2}}} - \frac{{Gm_r^2}}{{{d^2}}} = 0 \cr & \Rightarrow {\left( {\Delta e} \right)^2} = \frac{{Gm_r^2}}{K} \cr & \Rightarrow {\left( {\Delta e} \right)^2} = \frac{{\left( {6.67 \times {{10}^{ - 11}}} \right){{\left( {1.67 \times {{10}^{ - 27}}} \right)}^2}}}{{9 \times {{10}^9}}} \cr & \Rightarrow \Delta e = 1.437 \times {10^{ - 37}}C \cr} $$

Net force on charge placed at $$A$$ due to charges placed at $$C$$ and $$B$$

$$\eqalign{ & {F_A} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{\left( { + 4q} \right)\left( { - q} \right)}}{{{a^2}}} + \frac{1}{{4\pi {\varepsilon _0}}}\frac{{\left( { + 4q} \right)\left( { + 4q} \right)}}{{{{\left( {2a} \right)}^2}}} \cr & = - \frac{1}{{4\pi {\varepsilon _0}}}\frac{{4{q^2}}}{{{a^2}}} + \frac{1}{{4\pi {\varepsilon _0}}}\frac{{4{q^2}}}{{{a^2}}} = 0 \cr} $$
Similarly, $${F_B}$$ and $${F_C}$$ will be zero.
As net force on each charge is zero, therefore all the charges are in equilibrium. If we were to displace $$-q$$  to the right, (in figure), net force of attraction will be to the right which will displace it further. Therefore, equilibrium is unstable.

Let $$F$$ be the force between $$Q$$ and $$Q.$$ The force between $$q$$ and $$Q$$ should be attractive for net force on $$Q$$ to be zero. Let $$F'$$ be the force between $$Q$$ and $$q.$$ For equilibrium $$\sqrt 2 F' = - F$$
$$\eqalign{ & \sqrt 2 \times k\frac{{Qq}}{{{\ell ^2}}} = - k\frac{{{Q^2}}}{{{{\left( {\sqrt 2 \ell } \right)}^2}}} \cr & \Rightarrow \frac{Q}{q} = - 2\sqrt 2 \cr} $$

Electric potential due to charge $$Q$$ placed at the centre of the spherical shell at point $$P$$ is
$${V_1} = \frac{1}{{4\pi {\varepsilon _o}}}\frac{Q}{{\frac{R}{2}}} = \frac{1}{{4\pi {\varepsilon _o}}}\frac{{2Q}}{R}$$

Electric potential due to charge $$q$$ on the surface of the spherical shell at any point inside the shell is
$${V_2} = \frac{1}{{4\pi {\varepsilon _o}}}\frac{q}{R}$$
$$\therefore $$ The net electric potential at point $$P$$ is
$$V = {V_1} + {V_2} = \frac{1}{{4\pi {\varepsilon _o}}}\frac{{2Q}}{R} + \frac{1}{{4\pi {\varepsilon _o}}}\frac{q}{R}$$

$$\eqalign{ & eV = \frac{1}{2}m{v^2} \cr & \Rightarrow v = \sqrt {\frac{{2eV}}{m}} = \sqrt {\frac{{2 \times 1.6 \times {{10}^{ - 19}} \times 20}}{{9.1 \times {{10}^{ - 31}}}}} \cr & = 2.65 \times {10^6}m/s \cr} $$

Surface charge density $$\left( \sigma \right) = \frac{{{\text{Charge}}}}{{{\text{Surface area}}}}$$
$$\eqalign{ & {\text{So}}\,{\sigma _{{\text{inner}}}} = \frac{{ - 2Q}}{{4\pi {b^2}}} \cr & {\text{and}}\,{\sigma _{{\text{Outer}}}} = \frac{Q}{{4\pi {c^2}}} \cr} $$

Let two equal charges $$Q$$ each be held at $$A$$ and $$B,$$ where $$AB = 2x. C$$   is the centre of $$AB,$$  where charge $$q$$ is held.
Net force on $$q$$ is zero. So, $$q$$ is already in equilibrium.

For the three charges to be in equilibrium, net force on each charge must be zero.
Now, total force on $$Q$$ at $$B$$ is $$\frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{{Qq}}{{{x^2}}} + \frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{{QQ}}{{{{\left( {2x} \right)}^2}}} = 0$$
$$\eqalign{ & {\text{or}}\,\,\frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{{Qq}}{{{x^2}}} = - \frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{{{Q^2}}}{{4{x^2}}} \cr & {\text{or}}\,\,q = - \frac{Q}{4} \cr} $$