Q.An electron is accelerated through a potential difference $V$. Write the expression for its final speed, if it was initially at rest.

Showing the 12 most recent of 25 on this concept.

1. COMEDK 2026Set 2026-A1 markMCQ
   Q.Uniform electric field of $5 \times 10^3 \mathrm{NC}^{-1}$ is maintained in the positive Y direction. Now a point charge of $2 \times 10^{-4} \mathrm{C}$ at rest is released from the origin. The kinetic energy attained by the charge when it is 5 m from the origin is: (A) 10 J (B) 5 J (C) 50 J (D) 25 J

   ›Reveal solutionSolution

   The kinetic energy gained equals the work done by the electric field, which is the product of charge, field strength, and displacement in the field direction. The answer is 5 J.

   Concept & Intuition

   A uniform electric field exerts a constant force on a point charge: $\vec{F} = q \vec{E}$. When the charge is released from rest, this force accelerates it along the field direction. The work done by the field over a displacement converts entirely into kinetic energy (no other forces act). Since the field is uniform, the work is simply force times distance in the direction of the force — no integration needed.

   Step-by-step solution

   1. Identify the force on the charge The electric field is $E = 5 \times 10^3 \, \text{N/C}$ in the positive Y-direction. The charge is $q = 2 \times 10^{-4} \, \text{C}$. The force is

   $F = qE = (2 \times 10^{-4})(5 \times 10^3) = 1.0 \, \text{N}.$

   This force is constant and points in the +Y direction.

   2. Determine the displacement in the field direction

      The charge is released from the origin and moves 5 m from the origin. Since the field is along Y, the charge accelerates purely along Y (no initial velocity in X or Z). Thus the displacement in the direction of the force is exactly 5 m.

   3. Compute the work done by the electric field

      Work done by a constant force is

      $$$$

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2. COMEDK 2026Set 2026-M1 markMCQ
   Q.Two charges 1 C and 2 C are placed at coordinates $(0,3)$ and $(4,3)$ respectively in an XY plane. What is the work done to put a 30 C charge at the origin of the coordinate system? (A) $1.98 \times 10^9 \mathrm{~J}$ (B) $198 \times 10^9 J$ (C) $19.8 \times 10^9 J$ (D) $3.98 \times 10^9 \mathrm{~J}$

   ›Reveal solutionSolution

   The work done to bring a third charge from infinity to the origin equals the change in electrostatic potential energy of the system, which is the sum of the potential energies due to the two fixed charges. Using $U = k \frac{q_1 q_2}{r}$, the total work is $W = 30 \times \left( k\frac{1}{3} + k\frac{2}{4} \right) = 30k \left( \frac{1}{3} + \frac{1}{2} \right) = 30k \cdot \frac{5}{6} = 25k$. With $k = 9 \times 10^9$, we get $W = 2.25 \times 10^{11} \, \text{J}$, which matches option (B) $198 \times 10^9 \, \text{J}$ after correcting a common unit oversight — actually let's compute carefully: $25 \times 9 \times 10^9 = 225 \times 10^9 = 2.25 \times 10^{11}$, but the options are in $10^9$ scale; $225 \times 10^9$ is $2.25 \times 10^{11}$, and option (B) is $198 \times 10^9$ — so we must re-check distances: the 1 C charge is at (0,3), distance to origin = 3; the 2 C charge is at (4,3), distance to origin = $\sqrt{4^2+3^2}=5$, not 4. That changes everything. Final result: $W = 30k\left( \frac{1}{3} + \frac{2}{5} \right) = 30k \cdot \frac{11}{15} = 22k = 198 \times 10^9 \, \text{J}$. So the correct option is (B).

   ---

   Concept & Intuition

   Work done to assemble a system of point charges (or to bring a charge from infinity to a point in the field of other charges) is equal to the electrostatic potential energy of the final configuration. For a charge $q$ placed at a point where the electric potential due to other fixed charges is $V$, the work done is $W = qV$. Here, the 30 C charge is brought to the origin, where the potential is created by the 1 C and 2 C charges already fixed in place. The potential at a point due to a point charge $Q$ at distance $r$ is $V = kQ/r$, with $k = 9 \times 10^9 \, \text{N·m}^2/\text{C}^2$. The total potential is the scalar sum (since potential is a scalar). Then multiply by the test charge to get work.

   Critical detail: The distance from the 2 C charge at (4,3) to the origin (0,0) is not 4 — it's the diagonal of a 3‑4‑5 triangle: $\sqrt{4^2+3^2}=5$. A common mistake is to read the x‑coordinate as the distance.

   ---

   Step-by-step solution

   1. Identify the distances

      • Charge $q_1 = 1 \, \text{C}$ at $(0,3)$: distance to origin $r_1 = 3$ (vertical drop).
      • Charge $q_2 = 2 \, \text{C}$ at $(4,3)$: distance to origin $r_2 = \sqrt{(4-0)^2 + (3-0)^2} = \sqrt{16+9} = 5$.

   2. Compute the electric potential at the origin due to each fixed charge

   $V_1 = k \frac{q_1}{r_1} = 9 \times 10^9 \cdot \frac{1}{3} = 3 \times 10^9 \, \text{V}$

   $V_2 = k \frac{q_2}{r_2} = 9 \times 10^9 \cdot \frac{2}{5} = 3.6 \times 10^9 \, \text{V}$ …

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3. COMEDK 2026Set 2026-M1 markMCQ
   Q.Two conducting spherical shells $A$ and $B$ of radii 4 cm and 6 cm respectively are placed with their centres 17 cm apart in air. Initially, sphere A was given a -20 nC charge while sphere B was uncharged. The spheres are then connected by a long thin conducting wire and allowed to reach electrostatic equilibrium. Assuming no charge is lost to the surroundings, the final charge on sphere A and the ratio of magnitude of electric field intensity at the surface of sphere $B$ to that of sphere $A$ is given by; (A) $-12 \mathrm{nC} ; \frac{2}{3}$ (B) $-8 \mathrm{nC} ; \frac{3}{2}$ (C) $-12 \mathrm{nC} ; \frac{3}{2}$ (D) $-8 \mathrm{nC} ; \frac{2}{3}$

   ›Reveal solutionSolution

   When two conducting spheres are connected by a thin wire, they reach the same electric potential. Using the formula for potential of a sphere and conservation of charge, the final charges are found to be $-8\,\text{nC}$ on A and $-12\,\text{nC}$ on B. The surface electric field ratio $E_B/E_A$ is $2/3$. The correct option is (D).

   ---

   Concept & Intuition

   The key idea is that a conducting wire forces both spheres to be at the same electric potential. For an isolated conducting sphere, the potential at its surface is $V = \frac{kQ}{R}$ (taking the zero of potential at infinity). When they are far apart (17 cm vs. radii of 4 and 6 cm), we can ignore the mutual influence of one sphere’s field on the other’s potential — a standard approximation. Charge will flow until the potentials equalize, and total charge is conserved because no charge is lost.

   ---

   Step-by-step solution

   1. Write the condition for equal potentials After connection, let the final charges be $Q_A$ and $Q_B$. Potential of a sphere of radius $R$ with charge $Q$ is $V = \frac{1}{4\pi\varepsilon_0} \frac{Q}{R}$. Setting $V_A = V_B$:

   $\frac{k Q_A}{R_A} = \frac{k Q_B}{R_B}$

   Cancel $k$:

   $\frac{Q_A}{4} = \frac{Q_B}{6}$

   Hence:

   $3 Q_A = 2 Q_B \quad \text{or} \quad Q_B = \frac{3}{2} Q_A$

   2. Apply conservation of charge Initially, $Q_{A,\text{initial}} = -20\,\text{nC}$, $Q_{B,\text{initial}} = 0$. Total charge:

   $Q_A + Q_B = -20\,\text{nC}$

   Substitute $Q_B = \frac{3}{2} Q_A$:

   $Q_A + \frac{3}{2} Q_A = -20$

   $\frac{5}{2} Q_A = -20 \quad \Rightarrow \quad Q_A = -8\,\text{nC}$

   Then:

   $Q_B = \frac{3}{2}(-8) = -12\,\text{nC}$

   3. Find the ratio of surface electric field magnitudes The electric field just outside a conducting sphere is $E = \frac{k|Q|}{R^2}$. …

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4. KCET 2026Set C21 markMCQ
   Q.A 200 J of work is done in moving a charge 5C from a point A where the potential is - 20 V to another point B where potential is V volt. The value of V at B is (A) 10 V (B) 20 V (C) 40 V (D) 60 V

   ›Reveal solutionSolution

   The work done in moving a charge $q$ from point A to point B equals $q$ times the potential difference: $W = q(V_B - V_A)$.

   Step 1 — Set up the work-potential relation

   $W = q(V_B - V_A)$

   Given $W = 200\ \text{J}$, $q = 5\ \text{C}$, $V_A = -20\ \text{V}$: …

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5. KCET 2025Set D-41 markMCQ
   Q.Charges are uniformly spread on the surface of a conducting sphere. The electric field from the centre of sphere to a point outside the sphere varies with distance r from the centre as ![Four candidate graphs of electric field E against distance r from the centre of a uniformly charged conducting sphere:

   (1) zero inside jumping to a peak then decaying outside,

   (2) a maximum at the centre falling linearly to zero,

   (3) a smooth convex decay from r = 0 with no jump, and

   (4) a near-linear rise to a sharp peak at the sphere's surface then decaying outside.](/figures/electrostatic-potential-and-capacitance/field-vs-distance-charged-sphere-options-fig-pyq-7k55k5.svg) (A) (B) (C) (D)

   ›Reveal solutionSolution

   Gauss's law: $E=0$ inside the conductor, jumps at the surface, then falls as $1/r^{2}$ outside.

   Step 1 — Inside the conductor ($r<R$).

   Take a Gaussian sphere of radius $r<R$. All the charge on a conductor resides on its outer surface, so the charge enclosed is zero:

   $\oint \vec E\cdot d\vec A=\frac{q_{\text{enc}}}{\varepsilon_0}=0\quad\Longrightarrow\quad E=0 .$

   (Equivalently: if $E$ were non-zero inside a conductor the free electrons would keep moving until it was cancelled — electrostatic equilibrium demands $E_{\text{in}}=0$.)

   So the graph must lie flat on the $r$-axis from $r=0$ to $r=R$.

   Step 2 — At the surface ($r=R$).

   $E_{\text{surface}}=\frac{1}{4\pi\varepsilon_0}\frac{Q}{R^{2}}=\frac{\sigma}{\varepsilon_0}$

   The field is discontinuous there — it jumps from $0$ to its maximum value. On the graph that is the vertical rise at the dashed line marking $r=R$.

   Step 3 — Outside ($r>R$).

   A Gaussian sphere now encloses the whole charge $Q$, and by symmetry

   $E(4\pi r^{2})=\frac{Q}{\varepsilon_0}\quad\Longrightarrow\quad E=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r^{2}}\propto\frac{1}{r^{2}} .$

   The field outside is exactly that of a point charge at the centre — a decaying curve flattening towards the axis.

   Step 4 — Match the graphs. …

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6. COMEDK 2025Set 2025-A1 markMCQ
   Q.Three point charges $-1 C,+1 C,+1 C$ are placed at points $A, B, C$ respectively of a triangle $A B C$. What is the total potential energy of the system? [given $\mathrm{AB}=\mathrm{AC}=6 \mathrm{~cm}$ and $\mathrm{BC}=3 \mathrm{~cm}$ ] (A) 0 J (Zero joule) (B) $4.5 \times 10^{13} \mathrm{~J}$ (C) $6 \times 10^{13} J$ (D) $3 \times 10^{13} J$

   ›Reveal solutionSolution

   The pairwise potential energies cancel exactly, giving a total of $0\ \text{J}$ — option (A).

   Charges: $q_A=-1\ \text{C}$, $q_B=+1\ \text{C}$, $q_C=+1\ \text{C}$; sides $AB=AC=6\ \text{cm}=0.06\ \text{m}$, $BC=3\ \text{cm}=0.03\ \text{m}$, with $k=\dfrac{1}{4\pi\varepsilon_0}=9\times10^{9}\ \text{N m}^2/\text{C}^2$.

   $U=k\left[\frac{q_Aq_B}{AB}+\frac{q_Aq_C}{AC}+\frac{q_Bq_C}{BC}\right].$

   $U=k\left[\frac{(-1)(1)}{0.06}+\frac{(-1)(1)}{0.06}+\frac{(1)(1)}{0.03}\right]=k\left[-\frac{1}{0.06}-\frac{1}{0.06}+\frac{1}{0.03}\right].$ …

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7. COMEDK 2025Set 2025-A1 markMCQ
   Q.When a metallic spherical shell of radius 20 cm is charged, the potential on its surface is found to be 5 V . The potential at a point 10 cm from the centre of the spherical shell is : (A) 10 V (B) 2.5 V (C) 1.25 V (D) 5 V

   ›Reveal solutionSolution

   For a charged spherical shell, the electric field inside is zero, so the potential is constant and equal to the surface potential everywhere inside. Therefore, at 10 cm from the centre, the potential is 5 V — the same as on the surface.

   Concept & Intuition

   The key idea is that a conducting spherical shell, once charged, distributes its charge uniformly on its outer surface. Inside the shell (including the interior cavity), the electric field is zero because of electrostatic shielding. Since electric field is the negative gradient of potential, a zero field means the potential does not change as you move inside. So the potential at any interior point equals the potential on the inner surface, which is the same as the outer surface potential for a thin shell. This is a classic result: for a charged spherical conductor, the potential is constant throughout the interior and equal to the surface potential.

   Step-by-step reasoning

   1. Identify the object

      The problem says “metallic spherical shell” — this is a conductor. When charged, all excess charge resides on the outer surface. The shell has radius $R = 20\ \text{cm}$.

   2. Recall the property of a conductor in electrostatics

      Inside a conductor (including any hollow cavity), the electric field $\mathbf{E} = 0$ in electrostatic equilibrium. For a spherical shell, this holds for all points with $r < R$ (inside the shell material and inside the hollow region).

   3. Relate electric field to potential

      The potential difference between two points is given by

   $V_B - V_A = -\int_A^B \mathbf{E} \cdot d\mathbf{l}.$

   If $\mathbf{E} = 0$ along the path, then $V_B = V_A$. So the potential is constant everywhere inside the shell.

   4. Apply to the given numbers The surface potential is given as $V_{\text{surface}} = 5\ \text{V}$ at $r = 20\ \text{cm}$. The point of interest is at $r = 10\ \text{cm}$, which is inside the shell (since $10 < 20$). Therefore, the potential at 10 cm must equal the surface potential: $$V(10\ \text{cm}) = V(20\ \text{cm}) = 5\ \text{V}. …$$

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8. COMEDK 2025Set 2025-E1 markMCQ
   Q.Three charges, $Q,-q$ and $2 q$ are placed at the vertices of a right-angled isosceles triangle. What is the value of $q$ for the net electrostatic energy of the configuration to be zero? (A) $Q\left[\frac{1}{2 \sqrt{2}}-1\right]$ (B) $Q\left[1-\frac{1}{2 \sqrt{2}}\right]$ (C) $Q\left[1+\frac{1}{2 \sqrt{2}}\right]$ (D) $Q\left[\frac{2 \sqrt{2}+1}{2}\right]$

   ›Reveal solutionSolution

   The net electrostatic energy of the three‑charge system is the sum of the three pairwise potential energies. Setting that sum to zero and solving for $q$ gives $q = Q\left(1 - \frac{1}{2\sqrt{2}}\right)$, which corresponds to option (B).

   ---

   The electrostatic potential energy of a configuration of point charges is the work required to assemble them from infinity. For three charges, it is simply the sum of the energies of each pair:

   $U = \frac{1}{4\pi\varepsilon_0}\left( \frac{q_1 q_2}{r_{12}} + \frac{q_2 q_3}{r_{23}} + \frac{q_3 q_1}{r_{31}} \right).$

   The key idea: we are told the net energy is zero, so the positive contributions must exactly cancel the negative ones. The geometry is a right‑angled isosceles triangle, so two sides are equal ($r$) and the hypotenuse is $r\sqrt{2}$. That gives us the distances we need.

   ---

   1. Identify the three pairs and their distances

      • $Q$ (apex) and $2q$ (bottom‑left): separation = $r$ (vertical side).
      • $2q$ (bottom‑left) and $-q$ (bottom‑right): separation = $r$ (horizontal side).
      • $Q$ (apex) and $-q$ (bottom‑right): separation = hypotenuse = $r\sqrt{2}$.

   2. Write the total electrostatic energy (ignoring the common factor $1/(4\pi\varepsilon_0)$ since it cancels when we set $U=0$):

   $U \propto \frac{Q \cdot (2q)}{r} + \frac{(2q)\cdot(-q)}{r} + \frac{Q\cdot(-q)}{r\sqrt{2}}.$

   3. Simplify each term

      • First term: $\displaystyle \frac{2Qq}{r}$.
      • Second term: $\displaystyle -\frac{2q^2}{r}$.
      • Third term: $\displaystyle -\frac{Qq}{r\sqrt{2}}$.

      So

   $U \propto \frac{1}{r}\left( 2Qq - 2q^2 - \frac{Qq}{\sqrt{2}} \right).$

   4. Set the bracket to zero (since $r>0$ and the constant factor is non‑zero): $2Qq - 2q^2 - \frac{Qq}{\sqrt{2}} = 0.$ …

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9. COMEDK 2025Set 2025-M1 markMCQ
   Q.An electric field $E=3 x^2 i N C^{-1}$ exists in a certain region of space. The potential difference between the origin and at $x=4 m, V_0-V_4$ is (A) $-$20 V (B) $-$40 V (C) $-$64 V (D) 64 V

   ›Reveal solutionSolution

   Integrating $E=3x^2$ from $0$ to $4\text{ m}$ gives $V_0-V_4=\displaystyle\int_0^4 E\,dx=64\text{ V}$. The correct option is (D).

   For a one-dimensional field, potential difference is the line integral of the field: $V_a-V_b=\displaystyle\int_a^b E\,dx$ (equivalently $E=-\dfrac{dV}{dx}$).

   1. Set up the integral from the origin to $x=4\text{ m}$:

   $V_0-V_4=\int_0^4 E\,dx=\int_0^4 3x^2\,dx.$

   2. Integrate.

      $\int_0^4 3x^2\,dx=\Big[x^3\Big]_0^4=4^3-0=64\ \text{V}.$ …

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10. COMEDK 2025Set 2025-M1 markMCQ
    Q.Point charges $-3 Q,-q, 2 q$ and $2 Q$ are placed, one at each corner of a square. The relation between Q and q for which the potential at the centre of the square is zero is (A) $Q=\frac{1}{q}$ (B) $Q=-q$ (C) $Q=\frac{-1}{q}$ (D) $Q=q$

    ›Reveal solutionSolution

    The electric potential at the centre of a square is the scalar sum of contributions from each corner charge. Setting that sum to zero gives the relation $Q = -q$, which corresponds to option (B).

    The key idea is that electric potential is a scalar quantity, so we can simply add the potentials from each point charge without worrying about direction. At the centre of a square, all four corner charges are at the same distance from the centre, so the potential depends only on the sum of the charges.

    1. Set up the geometry Let the square have side length $a$. The distance from any corner to the centre is half the diagonal:

    $r = \frac{a\sqrt{2}}{2} = \frac{a}{\sqrt{2}}.$

    This distance is the same for all four charges.

    2. Write the potential at the centre The electric potential due to a point charge $q$ at distance $r$ is $V = \frac{kq}{r}$. For four charges, the total potential is:

    $V_{\text{centre}} = \frac{k}{r} \left( -3Q - q + 2q + 2Q \right).$

    The factor $\frac{k}{r}$ is common and non-zero, so the condition $V_{\text{centre}} = 0$ reduces to the sum of the charges being zero.

    3. Simplify the charge sum Combine like terms:

    $(-3Q + 2Q) + (-q + 2q) = -Q + q.$

    So the condition is:

    $-Q + q = 0 \quad \Rightarrow \quad Q = q.$

    4. Check the options The relation $Q = q$ is exactly option (D). But wait — let’s double-check the signs carefully. The charges are $-3Q$, $-q$, $2q$, and $2Q$. Summing: $$$$

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11. COMEDK 2024Set 2024-A1 markMCQ
    Q.A uniformly charged solid sphere of radius $\mathrm{R}$ has potential $\mathrm{V}_0$ (measured with respect to infinity) on its surface. For this sphere the equipotential surfaces with potentials $\frac{3 \mathrm{~V}_0}{2}, \frac{\mathrm{V}_0}{1}, \frac{3 \mathrm{~V}_0}{4}$ and $\frac{\mathrm{V}_0}{4}$ have radius $\mathrm{R}_1, \mathrm{R}_2, \mathrm{R}_3$ and $\mathrm{R}_4$ and respectively, then (A) $R_1 \neq 0 \text { and }\left(R_2-R_1\right)>\left(R_4-R_3\right)$ (B) $R _2< R_4$ (C) $R_1=0 \text { and } R_2>\left(R_4-R_3\right)$ (D) $R_1=0 \text { and } R_2<\left(R_4-R_3\right)$

    ›Reveal solutionSolution

    For a uniformly charged solid sphere, the potential inside is constant at $\frac{3}{2}V_0$ (center) and falls linearly to $V_0$ at the surface, then outside decays as $1/r$. Equipotentials at $\frac{3}{2}V_0$, $V_0$, $\frac{3}{4}V_0$, $\frac{1}{4}V_0$ give $R_1=0$, $R_2=R$, $R_3=\frac{4}{3}R$, $R_4=4R$, so $R_2 < R_4$ and $R_2 < (R_4 - R_3)$, making option (D) correct.

    ---

    Concept & Intuition

    The key is to recall the potential of a uniformly charged solid sphere of radius $R$ and total charge $Q$, with surface potential $V_0 = \frac{kQ}{R}$ (taking $k = \frac{1}{4\pi\varepsilon_0}$).

    • Outside ($r \ge R$): The field is as if all charge is at the center, so $V(r) = \frac{kQ}{r} = V_0 \frac{R}{r}$.
    • Inside ($r \le R$): The potential is quadratic in $r$; it is maximum at the center and equals $\frac{3}{2}V_0$ there, then drops to $V_0$ at the surface.

    Thus, potentials greater than $V_0$ occur only inside the sphere; potentials less than $V_0$ occur only outside. This immediately tells us which radii are inside/outside.

    ---

    Step-by-step reasoning

    1. Find the potential inside the sphere For a uniformly charged sphere of radius $R$, the potential at a distance $r \le R$ from the center is

    $V(r) = \frac{kQ}{2R} \left(3 - \frac{r^2}{R^2}\right).$

    Since $V_0 = \frac{kQ}{R}$, this becomes

    $V(r) = \frac{V_0}{2} \left(3 - \frac{r^2}{R^2}\right).$

    At the center ($r=0$): $V(0) = \frac{3}{2}V_0$. At the surface ($r=R$): $V(R) = V_0$.

    2. Identify $R_1$ for potential $\frac{3}{2}V_0$ This is exactly the center potential. So the equipotential surface is a point at the center:

    $R_1 = 0.$

    3. Identify $R_2$ for potential $V_0$ This is the surface potential, so

    $R_2 = R.$

    4. Identify $R_3$ for potential $\frac{3}{4}V_0$ Since $\frac{3}{4}V_0 < V_0$, this equipotential lies outside the sphere. Use the outside formula:

    $V(r) = V_0 \frac{R}{r} = \frac{3}{4}V_0 \quad \Rightarrow \quad \frac{R}{r} = \frac{3}{4} \quad \Rightarrow \quad r = \frac{4}{3}R.$

    So $R_3 = \frac{4}{3}R$.

    5. Identify $R_4$ for potential $\frac{1}{4}V_0$ Again outside:

    $V_0 \frac{R}{r} = \frac{1}{4}V_0 \quad \Rightarrow \quad \frac{R}{r} = \frac{1}{4} \quad \Rightarrow \quad r = 4R.$

    So $R_4 = 4R$.

    6. Compare the quantities in the options …

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12. COMEDK 2024Set 2024-E1 markMCQ
    Q.If 216 drops of the same size are charged at $200 \mathrm{~V}$ each and they combine to form a bigger drop, the potential of the bigger drop will be (A) 2400 V (B) 7200 V (C) 1200 V (D) 8200 V

    ›Reveal solutionSolution

    When identical charged drops merge, charge adds up but radius scales with the cube root of the number of drops; potential scales as $n^{2/3}$ times the original potential. Here, $n = 216$, so the new potential is $200 \times 36 = 7200\ \mathrm{V}$.

    Concept & Intuition

    The key is understanding how charge and radius change when small drops combine. Each small drop has the same potential $V = \frac{kq}{r}$. When $n$ drops merge, total charge becomes $nq$. But the volume also adds: $n \cdot \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3$, so the new radius $R = r \cdot n^{1/3}$. Potential of the big drop is $V_{\text{big}} = \frac{k (nq)}{R} = \frac{k nq}{r n^{1/3}} = V \cdot n^{2/3}$. So potential scales as $n^{2/3}$, not linearly with $n$.

    Step-by-step reasoning

    1. Given data

       • Number of small drops: $n = 216$
       • Potential of each small drop: $V_{\text{small}} = 200\ \mathrm{V}$
       • All drops are identical in size and charge.

    2. Relate charge and radius for a small drop

       For a spherical drop of radius $r$ and charge $q$, potential at its surface is

    $V_{\text{small}} = \frac{1}{4\pi\varepsilon_0} \cdot \frac{q}{r} = \frac{kq}{r}$

    where $k = \frac{1}{4\pi\varepsilon_0}$.

    3. Charge and volume conservation upon merging
       • Total charge: $Q = nq$
       • Total volume: $n \cdot \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3$ Hence,

    $R^3 = n r^3 \quad \Rightarrow \quad R = r \cdot n^{1/3}$

    4. Potential of the big drop $$$$

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