Question 1

   Question: Two identical charged conducting spheres $A$ and $B$ have their centres separated by a certain distance. Charge on each sphere is $q$ and the force of repulsion between them is $F$. A third identical uncharged conducting sphere is brought in contact with sphere $A$ first and then with $B$ and finally removed from both. New force of repulsion between spheres $A$ and $B$ (Radii of $A$ and $B$ are negligible compared to the distance of separation so that for calculating force between them they can be considered as point charges) is best given as:

   Options: 

    A. $\frac{F}{2}$

    B. $\frac{3F}{8}$

    C. $\frac{3F}{5}$

    D. $\frac{2F}{3}$

   Correct Answer: B

   Year: NEET 2025

   Solution: 

    $F = \frac{K q q}{r^2}$

    $F' = \frac{\frac{K q}{2} \frac{3 q}{4}}{r^2}$

    $F' = \frac{3 F}{8}$

   Step Solution:

    1.  The initial force between spheres A and B with charge $q$ is $F = \frac{kq^2}{r^2}$.

    2.  When uncharged sphere C touches A, the charge is shared equally: $q_A = \frac{q + 0}{2} = \frac{q}{2}$.

    3.  When sphere C (now with charge $\frac{q}{2}$) touches B (charge $q$), the new charge on B is $q_B = \frac{\frac{q}{2} + q}{2} = \frac{3q}{4}$.

    4.  The new force $F'$ is calculated using the new charges: $F' = \frac{k(\frac{q}{2})(\frac{3q}{4})}{r^2}$.

    5.  Simplifying the math: $F' = \frac{3}{8} \left( \frac{kq^2}{r^2} \right) = \frac{3F}{8}$.

   Difficulty level: Medium

   Concept Name: Coulomb's Law and Charge Distribution on Conductors

   Short cut solution: The final force is the product of the fractional charges: $\frac{1}{2} \times \frac{3}{4} = \frac{3}{8}$ of the original force.

Question 38

   Question: Two point charges A and B, having charges $+Q$ and $-Q$ respectively, are placed at certain distance apart and force acting between them is $F$. If $25\%$ charge of A is transferred to B, then force between the charges becomes:

   Options: 

    A. $\frac{4F}{3}$

    B. $F$

    C. $\frac{9F}{16}$

    D. $\frac{16F}{9}$

   Correct Answer: C

   Year: NEET 2019

   Solution: 

    In case I: $F = - \frac{1}{4 \pi \varepsilon_0} \frac{Q^2}{r^2}$

    In Case II: $Q_A = Q - \frac{Q}{4}, Q_B = - Q + \frac{Q}{4}$

    $\therefore F' = \frac{1}{4 \pi \varepsilon_0} \frac{(Q - \frac{Q}{4}) (- Q + \frac{Q}{4})}{r^2} = - \frac{1}{4 \pi \varepsilon_0} \frac{9}{16} \frac{Q^2}{r^2} = \frac{9}{16} F$

   Step Solution:

    1.  Initial force $F$ between $+Q$ and $-Q$ is proportional to $|Q \times -Q| = Q^2$.

    2.  $25\%$ of charge A is $\frac{Q}{4}$. Transferring this to B changes the charges.

    3.  New charge on A: $Q_A = Q - \frac{Q}{4} = \frac{3Q}{4}$.

    4.  New charge on B: $Q_B = -Q + \frac{Q}{4} = -\frac{3Q}{4}$.

    5.  New force $F' \propto |\frac{3Q}{4} \times -\frac{3Q}{4}| = \frac{9}{16} Q^2$, which means $F' = \frac{9}{16} F$.

   Difficulty level: Easy

   Concept Name: Coulomb's Law and Charge Transfer

   Short cut solution: After transferring $25\%$ ($1/4$), the remaining charge on each is $3/4$ of the original magnitude. The force scale is $(3/4)^2 = 9/16$.

Question 43

   Question: The electrostatic force between the metal plates of an isolated parallel plate capacitor $C$ having a charge $Q$ and area $A$, is:

   Options: 

    A. Independent of the distance between the plates

    B. Linearly proportional to the distance between the plates

    C. Proportional to the square root of the distance between the plates

    D. Inversely proportional to the distance between the plates

   Correct Answer: A

   Year: NEET 2018

   Solution: For isolated capacitor, charge $Q =$ constant. Electrostatic force, $F_{\text{plate}} = \frac{Q^2}{2 A \varepsilon_0}$.

   Step Solution:

    1.  In an isolated capacitor, the charge $Q$ remains constant.

    2.  The electric field produced by one plate is $E = \frac{\sigma}{2\epsilon_0}$. [Not explicitly in source, but derived from the force formula provided].

    3.  The force on the second plate is $F = Q \times E_{\text{other plate}}$. [Not explicitly in source].

    4.  Substituting $\sigma = \frac{Q}{A}$, the force formula is $F = \frac{Q^2}{2A\epsilon_0}$.

    5.  Since the formula for $F$ depends only on $Q, A,$ and $\epsilon_0$, it is independent of the distance $d$.

   Difficulty level: Medium

   Concept Name: Electrostatic Force on Capacitor Plates

   Short cut solution: The electric field of a large charged plate is uniform (constant) and does not depend on distance; therefore, the force $F = QE$ is also constant.

Question 46

   Question: Suppose the charge of a proton and an electron differ slightly. One of them is $-e$, the other is $(e + \Delta e)$. If the net of electrostatic force and gravitational force between two hydrogen atoms placed at a distance $d$ (much greater than atomic size) apart is zero, then $\Delta e$ is of the order of [given mass of hydrogen $m_h = 1.67 \times 10^{-27} \text{ kg}$]

   Options: 

    A. $10^{-23} \text{ C}$

    B. $10^{-37} \text{ C}$

    C. $10^{-47} \text{ C}$

    D. $10^{-20} \text{ C}$

   Correct Answer: B

   Year: 2017 NEET

   Solution: A hydrogen atom consists of an electron and a proton. ∴ Charge on one hydrogen atom $= q_e + q_p = -e + (e + \Delta e) = \Delta e$. Since a hydrogen atom carries a net charge $\Delta e$, $F_e = \frac{1}{4\pi\varepsilon_0} \frac{(\Delta e)^2}{d^2}$ will act between two hydrogen atoms. The gravitational force between two hydrogen atoms is given as $F_g = \frac{G m_h m_h}{d^2}$. Since the net force on the system is zero, $F_e = F_g$. Using eqns. (i) and (ii), we get $\frac{(\Delta e)^2}{4\pi\varepsilon_0 d^2} = \frac{G m_h^2}{d^2} \Rightarrow (\Delta e)^2 = 4\pi\varepsilon_0 G m_h^2 = 6.67 \times 10^{-11} \times \frac{(1.67 \times 10^{-27})^2}{(9 \times 10^9)}$. $\Delta e \approx 10^{-37} \text{ C}$.

   Step Solution:

    1.  Calculate net charge on one hydrogen atom: $q = q_p + q_e = (e + \Delta e) - e = \Delta e$.

    2.  Write the electrostatic force equation: $F_e = \frac{1}{4\pi\varepsilon_0} \frac{(\Delta e)^2}{d^2}$.

    3.  Write the gravitational force equation: $F_g = G \frac{m_h^2}{d^2}$.

    4.  Equate forces for zero net force: $\frac{(\Delta e)^2}{4\pi\varepsilon_0 d^2} = \frac{G m_h^2}{d^2} \implies (\Delta e)^2 = 4\pi\varepsilon_0 G m_h^2$.

    5.  Substitute values and find order: $(\Delta e)^2 \approx \frac{6.67 \times 10^{-11} \times (1.67 \times 10^{-27})^2}{9 \times 10^9} \approx 2 \times 10^{-74}$. Taking the square root, $\Delta e \approx 10^{-37} \text{ C}$.

   Difficulty level: Hard

   Concept Name: Balancing Gravitational and Electrostatic Forces

   Short cut solution: Equate $k(\Delta e)^2 \approx G m^2$. Rearrange to $\Delta e \approx m \sqrt{G/k}$. Using orders of magnitude: $10^{-27} \times \sqrt{10^{-11} / 10^9} = 10^{-27} \times 10^{-10} = 10^{-37}$.

Question 49

   Question: Two identical charged spheres suspended from a common point by two massless strings of lengths $l$, are initially at a distance $d$ $(d << l)$ apart because of their mutual repulsion. The charges begin to leak from both the spheres at a constant rate. As a result, the spheres approach each other with a velocity $v$. Then $v$ varies as a function of the distance $x$ between the spheres, as:

   Options: 

    A. $v \propto x^{-1/2}$

    B. $v \propto x^{-1}$

    C. $v \propto x^{1/2}$

    D. $v \propto x$

   Correct Answer: A

   Year: 2016 NEET Phase-I

   Solution: From figure, $T \cos\theta = mg$, $T \sin\theta = \frac{kq^2}{x^2}$. $\tan\theta = \frac{kq^2}{x^2 mg}$. Since $\theta$ is small, $\tan\theta \approx \sin\theta = \frac{x}{2l}$. $\therefore \frac{x}{2l} = \frac{kq^2}{x^2 mg} \Rightarrow q^2 = x^3 \frac{mg}{2lk}$ or $q \propto x^{3/2} \Rightarrow \frac{dq}{dt} \propto \frac{3}{2} \sqrt{x} \frac{dx}{dt} = \frac{3}{2} \sqrt{x} v$. Since $\frac{dq}{dt} = \text{constant}$, $\therefore v \propto \frac{1}{\sqrt{x}}$.

   Step Solution:

    1.  Equate forces at equilibrium: $T \sin\theta = \frac{kq^2}{x^2}$ and $T \cos\theta = mg$. Dividing gives $\tan\theta = \frac{kq^2}{x^2 mg}$.

    2.  Apply small angle approximation: $\tan\theta \approx \sin\theta = \frac{x/2}{l} = \frac{x}{2l}$.

    3.  Combine equations: $\frac{x}{2l} = \frac{kq^2}{x^2 mg} \implies q^2 \propto x^3$, which means $q \propto x^{3/2}$.

    4.  Differentiate with respect to time $t$: $\frac{dq}{dt} \propto \frac{d}{dt}(x^{3/2}) = \frac{3}{2} x^{1/2} \frac{dx}{dt}$.

    5.  Set $\frac{dq}{dt}$ as constant and $\frac{dx}{dt} = v$: $\text{Constant} \propto x^{1/2} v \implies v \propto x^{-1/2}$.

   Difficulty level: Hard

   Concept Name: Electrostatic Equilibrium and Rate of Change

   Short cut solution: From $x^3 \propto q^2$, if $dq/dt$ is constant, then $q \propto t$. Thus $x^3 \propto t^2 \implies x \propto t^{2/3}$. Velocity $v \propto t^{-1/3}$. Substituting $t \propto x^{3/2}$, we get $v \propto (x^{3/2})^{-1/3} = x^{-1/2}$.

Question 61

   Question: A charge $q$ is placed at the centre of the line joining two equal charges $Q$. The system of the three charges will be in equilibrium if $q$ is equal to:

   Options: 

    A. $-Q/4$

    B. $Q/4$

    C. $-Q/2$

    D. $Q/2$

   Correct Answer: A

   Year: KN NEET 2013

   Solution: Let two equal charges $Q$ each placed at points $A$ and $B$ at a distance $r$ apart. $C$ is the centre of $AB$ where charge $q$ is placed. For equilibrium, net force on charge $Q = 0$. $\therefore \frac{1}{4\pi\varepsilon_0} \frac{QQ}{r^2} + \frac{1}{4\pi\varepsilon_0} \frac{Qq}{(r/2)^2} = 0$. $\frac{1}{4\pi\varepsilon_0} \frac{Q^2}{r^2} = -\frac{1}{4\pi\varepsilon_0} \frac{4Qq}{r^2}$ or $Q = -4q$ or $q = -Q/4$.

   Step Solution:

    1.  The central charge $q$ is already in equilibrium because it is equidistant from two identical charges $Q$.

    2.  For the entire system to be in equilibrium, the net force on one of the outer charges ($Q$) must be zero.

    3.  The force on $Q$ from the other $Q$ (distance $r$) is $F_1 = \frac{kQ^2}{r^2}$.

    4.  The force on $Q$ from the central charge $q$ (distance $r/2$) is $F_2 = \frac{kqQ}{(r/2)^2} = \frac{4kqQ}{r^2}$.

    5.  Set $F_1 + F_2 = 0 \implies \frac{kQ^2}{r^2} + \frac{4kqQ}{r^2} = 0 \implies Q + 4q = 0$, so $q = -\frac{Q}{4}$.

   Difficulty level: Medium

   Concept Name: Electrostatic Equilibrium of a System

   Short cut solution: To balance the repulsion between two $+Q$ charges, the central charge must be negative. The distance factor is $(1/2)^2 = 1/4$, so the charge magnitude must be $1/4$ to cancel the force: $q = -Q/4$.

Question 62

   Question: Two pith balls carrying equal charges are suspended from a common point by strings of equal length, the equilibrium separation between them is $r$. Now the strings are rigidly clamped at half the height. The equilibrium separation between the balls now become:

   Options: 

    A. $(\frac{2r}{\sqrt{3}})$

    B. $(\frac{2r}{3})$

    C. $(\frac{1}{\sqrt{2}})^2$

    D. $\frac{r}{\sqrt{2}}$

   Correct Answer: D

   Year: 2013 NEET

   Solution: Let the length of the strings be $L$ and mass of the ball be $m$ and charge be $q$. At equilibrium, $\Sigma F_x = 0$ and $\Sigma F_y = 0$. $T \sin \theta = mg$ and $T \cos \theta = F_e \implies T \cos \theta = \frac{Kq^2}{r^2}$. Dividing gives $\tan \theta = \frac{kq^2}{r^2 mg}$. As $\tan \theta = \frac{r/2}{y}$, then $r^2 = C \times \frac{2y}{r} \Rightarrow r \propto (y)^{1/3}$. Thus $\frac{r'}{r} = \frac{(y')^{1/3}}{y^{1/3}}$. Since $y' = y/2$, $r' = \frac{r}{\sqrt{2}}$.

   Step Solution:

    1.  Establish equilibrium for the first case: $\tan \theta_1 = \frac{F_e}{mg} = \frac{kq^2}{r^2 mg}$.

    2.  Use the geometry of the setup where $\tan \theta_1 \approx \frac{r/2}{y} = \frac{r}{2y}$.

    3.  Equate the two expressions to find the relationship between distance $r$ and height $y$: $\frac{r}{2y} = \frac{kq^2}{r^2 mg} \implies r^3 = \frac{2ykq^2}{mg}$.

    4.  Identify the proportionality $r^3 \propto y$, which means $r \propto y^{1/3}$.

    5.  Substitute the new height $y' = y/2$ into the ratio: $\frac{r'}{r} = (\frac{y/2}{y})^{1/3} = (\frac{1}{2})^{1/3}$, resulting in $r' = \frac{r}{\sqrt{2}}$.

   Difficulty level: Hard

   Concept Name: Electrostatic Equilibrium of Suspended Charges

   Short cut solution: Use the derived relation $r \propto y^{1/3}$. When $y$ is halved, the new separation $r'$ is $r \times (\frac{1}{2})^{1/3} = \frac{r}{\sqrt{2}}$.

Question 72

   Question: Two positive ions, each carrying a charge $q$, are separated by a distance $d$. If $F$ is the force of repulsion between the ions, the number of electrons missing from each ion will be ($e$ being the charge on an electron):

   Options: 

    A. $\frac{4\pi\varepsilon_0 F d^2}{e^2}$

    B. $\sqrt{\frac{4\pi\varepsilon_0 F e^2}{d^2}}$

    C. $\sqrt{\frac{4\pi\varepsilon_0 F d^2}{e^2}}$

    D. $\frac{4\pi\varepsilon_0 F d^2}{q^2}$

   Correct Answer: C

   Year: 2010

   Solution: According to Coulomb’s law, $F = \frac{1}{4\pi\varepsilon_0} \frac{q^2}{d^2} \implies q^2 = 4\pi\varepsilon_0 F d^2$. Thus $q = \sqrt{4\pi\varepsilon_0 F d^2}$. Since $q = ne$, where $n$ is the number of missing electrons, $n = \frac{q}{e} = \sqrt{\frac{4\pi\varepsilon_0 F d^2}{e^2}}$.

   Step Solution:

    1.  Write the formula for Coulombic force: $F = \frac{1}{4\pi\varepsilon_0} \frac{q^2}{d^2}$.

    2.  Rearrange the formula to solve for the square of the charge: $q^2 = 4\pi\varepsilon_0 F d^2$.

    3.  Calculate the charge $q$ by taking the square root: $q = \sqrt{4\pi\varepsilon_0 F d^2}$.

    4.  Apply the principle of quantization of charge: $q = ne$.

    5.  Solve for $n$ by dividing the charge by the electronic charge: $n = \frac{\sqrt{4\pi\varepsilon_0 F d^2}}{e} = \sqrt{\frac{4\pi\varepsilon_0 F d^2}{e^2}}$.

   Difficulty level: Easy

   Concept Name: Coulomb's Law and Quantization of Charge

   Short cut solution: From $F = \frac{kq^2}{d^2}$, we have $q \propto \sqrt{F}d$. Since $n = q/e$, the answer must contain $\frac{d\sqrt{F}}{e}$.

Question 108

   Question: When air is replaced by a dielectric medium of constant $K$, the maximum force of attraction between two charges separated by a distance:

   Options: 

    A. increases $K$ times

    B. remains unchanged

    C. decreases $K$ times

    D. increases $K^{-1}$ times

   Correct Answer: C

   Year: 1999

   Solution: $F_m = \frac{F_0}{K}$ i.e., decreases $K$ times.

   Step Solution:

    1.  Recall the force between two charges in a vacuum: $F_0 = \frac{1}{4\pi\varepsilon_0} \frac{q_1q_2}{r^2}$.

    2.  Recall the force in a medium with dielectric constant $K$: $F_m = \frac{1}{4\pi\varepsilon} \frac{q_1q_2}{r^2}$.

    3.  Note that the permittivity of the medium is $\varepsilon = K\varepsilon_0$.

    4.  Substitute $\varepsilon$ into the force equation: $F_m = \frac{1}{4\pi(K\varepsilon_0)} \frac{q_1q_2}{r^2} = \frac{1}{K} (\frac{1}{4\pi\varepsilon_0} \frac{q_1q_2}{r^2})$.

    5.  Conclude that the new force is the original force divided by $K$: $F_m = \frac{F_0}{K}$.

   Difficulty level: Easy

   Concept Name: Effect of Dielectric Medium on Electrostatic Force

   Short cut solution: The electrostatic force is inversely proportional to the dielectric constant of the medium ($F \propto 1/K$).

Question 118

   Question: A charge $q$ is placed at the centre of the line joining two exactly equal positive charges $Q$. The system of three charges will be in equilibrium, if $q$ is equal to

   Options: 

    A. $-Q$

    B. $-Q/2$ (Inferred from context of similar problems in sources)

    C. $-Q/4$

    D. $+Q$

    (Note: The source provides the answer label as 'C' and the mathematical result as $-Q/4$)

   Correct Answer: C ($-Q/4$)

   Year: 1995

   Solution: For equilibrium of charge $Q$, the force of repulsion due to similar charges $Q$ should be balanced by the force of attraction due to charge $q$ and $Q$.

    $\frac{1}{4 \pi \varepsilon_{0}} \times \frac{Qq}{(r/2)^{2}} + \frac{1}{4 \pi \varepsilon_{0}} \times \frac{Q^{2}}{r^{2}} = 0$

    $4 \times \frac{Q}{r^{2}} q = -\frac{Q^{2}}{r^{2}}$ or $4q = -Q$ or $q = -Q/4$.

   Step Solution:

    1.  Place charges $+Q$ at $x = 0$ and $x = r$. Place $q$ at $x = r/2$.

    2.  The central charge $q$ is already in equilibrium due to symmetry.

    3.  For the system to be in equilibrium, the net force on an outer charge ($Q$) must be zero.

    4.  The force equation on $Q$ is: $\frac{kQ^2}{r^2} + \frac{kqQ}{(r/2)^2} = 0$.

    5.  Simplify: $\frac{kQ^2}{r^2} + \frac{4kqQ}{r^2} = 0 \implies Q + 4q = 0 \implies q = -Q/4$.

   Difficulty level: Medium

   Concept Name: Electrostatic Equilibrium of a System

   Short cut solution: For three collinear charges to be in equilibrium, the central charge must be negative and its magnitude must be $q = \frac{Q}{(1+1)^2} = Q/4$.

Question 123

   Question: Point charges $+4q, -q$ and $+4q$ are kept on the X-axis at point $X = 0, X = a$ and $x = 2a$ respectively. Then

   Options: 

    A. only $-q$ is in stable equilibrium

    B. all the charges are in stable equilibrium

    C. all of the charges are in unstable equilibrium

    D. none of the charges is in equilibrium

   Correct Answer: C

   Year: 1988

   Solution: Net force on each of the charge due to the other charges is zero. However, disturbance in any direction other than along the line on which the charges lie, will not make the charges return.

   Step Solution:

    1.  Calculate force on $-q$ at $x=a$: $F = \frac{k(4q)(-q)}{a^2} + \frac{k(4q)(-q)}{a^2} = -\frac{4kq^2}{a^2} + \frac{4kq^2}{a^2} = 0$.

    2.  Calculate force on $+4q$ at $x=0$: $F = \frac{k(4q)(-q)}{a^2} + \frac{k(4q)(4q)}{(2a)^2} = -\frac{4kq^2}{a^2} + \frac{16kq^2}{4a^2} = 0$.

    3.  Since net force on all charges is zero, the system is in equilibrium.

    4.  Analyze stability: If a charge is displaced perpendicular to the x-axis, the resulting forces will pull it further away rather than restoring it to the center.

    5.  Conclusion: Because it cannot return to its original position from all types of displacement, the equilibrium is unstable.

   Difficulty level: Hard

   Concept Name: Stability of Electrostatic Equilibrium

   Short cut solution: According to Earnshaw's Theorem, a collection of point charges cannot be maintained in a stable stationary equilibrium configuration solely by the electrostatic interaction of the charges. Thus, it must be unstable.