Q.Two point charges, each equal to $-q$, are held fixed on a straight line, separated by a distance $2d$ (so each is a distance $d$ from the mid-point of the line). A third charge $+q$ of mass $m$ is placed at the mid-point and is then displaced by a small distance $x$ (with $x \ll d$) in the direction perpendicular to the line joining the two fixed charges. Show that the charge $+q$ executes simple harmonic motion, and that its time period is $T = \left[\dfrac{8\pi^3 \varepsilon_0 m d^3}{q^2}\right]^{1/2}$.

Showing the 12 most recent of 22 on this concept.

1. COMEDK 2026Set 2026-A1 markMCQ
   Q.Two point charges $P=+25 \mu C$ and $Q=-16 \mu C$ are placed 5 cm apart. Find the position of the point at which the resultant electric field is zero: (A) 1 cm from $Q$ and 4 cm from $P$ on the dipole axis (B) 2.5 cm from $Q$ and 2.5 cm from $P$ on the dipole axis (C) 20 cm from $Q$ and 25 cm from $P$ on the dipole axis (D) 25 cm from $Q$ and 20 cm from $P$ on the dipole axis

   ›Reveal solutionSolution

   For two opposite charges, the net electric field is zero outside the smaller charge on the line joining them. Solving $E_P = E_Q$ gives distances 20 cm from Q and 25 cm from P, so option (C) is correct.

   Concept & Intuition

   The electric field due to a point charge is $E = k|q|/r^2$, directed away from a positive charge and toward a negative charge. Here we have $+25\,\mu\text{C}$ and $-16\,\mu\text{C}$. Because the charges are opposite, the fields point in opposite directions between the charges, but they can never cancel there because the magnitudes are different and the directions are opposite — the net field is nonzero.

   The only place where the fields can cancel is outside the smaller charge (the $-16\,\mu\text{C}$), because there the fields from both charges point in opposite directions (away from the positive, toward the negative) and the distances can be adjusted to make magnitudes equal.

   We set up the condition $|E_P| = |E_Q|$ and solve for the distances.

   Step-by-step solution

   1. Define the geometry

      Let the charges be on the x‑axis: $P$ at $x=0$ with $+25\,\mu\text{C}$, $Q$ at $x=5$ cm with $-16\,\mu\text{C}$.

      Let the zero‑field point be at distance $x$ from $P$ (so $x$ cm from $P$).

      If the point is to the right of $Q$, then distance from $P$ is $x$ and from $Q$ is $x-5$ (both in cm).

   2. Write the field magnitudes

   $E_P = \frac{k \cdot 25}{(x)^2}, \qquad E_Q = \frac{k \cdot 16}{(x-5)^2}$

   (We drop the sign because we only care about magnitude; direction is handled by the geometry.)

   3. Set magnitudes equal for cancellation

   $\frac{25}{x^2} = \frac{16}{(x-5)^2}$

   Take square roots (positive distances):

   $\frac{5}{x} = \frac{4}{x-5}$

   4. Solve for $x$ Cross‑multiply:

   $5(x-5) = 4x \quad\Rightarrow\quad 5x - 25 = 4x \quad\Rightarrow\quad x = 25\ \text{cm}$

   So the point is 25 cm from $P$. …

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2. COMEDK 2026Set 2026-M1 markMCQ
   Q.Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle of $45^{\circ}$ with each other. When the system is immersed in a liquid of relative density 0.8 , the angle between the strings remains unchanged. If the density of the material of the sphere is $2.4 \mathrm{gcm}^{-3}$, what is the dielectric constant of the liquid? (A) 1.6 (B) 1.5 (C) 0.5 (D) 2

   ›Reveal solutionSolution

   The key idea is that the angle remains the same because the electric force is reduced by the dielectric constant exactly as much as the effective weight is reduced by buoyancy. Solving the equilibrium condition gives the dielectric constant as 1.5.

   We have two identical charged spheres hanging from strings of equal length, making an angle of $45^\circ$ with each other. That means each string makes an angle of $22.5^\circ$ with the vertical. When immersed in a liquid, the angle stays the same. This tells us that the ratio of the electric repulsion to the effective weight (weight minus buoyancy) must be the same in air and in the liquid.

   Concept and intuition:

   In air, the spheres repel each other with a force $F_e = \frac{k q^2}{r^2}$ (where $k = 1/(4\pi\varepsilon_0)$). In the liquid, the electric force is reduced by the dielectric constant $\kappa$: $F_e' = \frac{k q^2}{\kappa r^2}$. Meanwhile, the weight of each sphere is $mg$, but in the liquid there is an upward buoyant force $F_b = \rho_{\text{liquid}} V g$. The effective weight becomes $mg - F_b = (\rho_{\text{sphere}} - \rho_{\text{liquid}}) V g$. Since the angle doesn’t change, the ratio of electric force to effective weight must be the same in both cases. That gives us an equation to solve for $\kappa$.

   Let’s work through it step by step.

   1. Set up the equilibrium in air. Each sphere has mass $m = \rho_s V$, where $\rho_s = 2.4 \, \text{g/cm}^3$ and $V$ is its volume. The tension in the string has vertical component $T \cos\theta = mg$ and horizontal component $T \sin\theta = F_e$, where $\theta = 22.5^\circ$ (half the angle between the strings). Dividing:

   $\tan\theta = \frac{F_e}{mg}.$

   2. Set up the equilibrium in the liquid. The effective weight is $mg - F_b = (\rho_s - \rho_l) V g$, with $\rho_l = 0.8 \, \text{g/cm}^3$ (relative density 0.8 means density 0.8 g/cm³). The electric force is now $F_e' = \frac{F_e}{\kappa}$ because the medium reduces the Coulomb force by the dielectric constant. The angle is the same, so:

   $\tan\theta = \frac{F_e'}{(\rho_s - \rho_l) V g}.$

   3. Equate the two expressions for $\tan\theta$. From air: $\tan\theta = \frac{F_e}{\rho_s V g}$. From liquid: $\tan\theta = \frac{F_e / \kappa}{(\rho_s - \rho_l) V g}$. …

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3. COMEDK 2026Set 2026-M1 markMCQ
   Q.Point charge $\sqrt{2} C, \sqrt{2} C$, and $-2 C$ are placed at the three vertices of a right-angled triangle in air. [as shown in the figure below] What is the electric field at a point $P$ on the hypotenuse that is equidistant from all three charges. Given distances $X P=Y P=Z P=0.5 \mathrm{~m}$ (A) $0 \cdot 72 \times 10^{10} N C^{-1} \quad$ along YP (B) $7.2 \times 10^{10} N C^{-1} \quad$ along PY (C) $0.72 \times 10^9 \mathrm{NC}^{-1} \quad$ along YP (D) $7.2 \times 10^9 \mathrm{NC}^{-1} \quad$ along PY

   ›Reveal solutionSolution

   The two equal $\sqrt{2}\,\text{C}$ charges sit at the ends of the hypotenuse, collinear through the midpoint $P$; their fields cancel there. Only the $-2\,\text{C}$ charge remains, giving $7.2\times10^{10}\,\text{N C}^{-1}$.

   Geometry

   $P$ is equidistant from all three vertices of a right-angled triangle, so it is the circumcentre — the midpoint of the hypotenuse — with $XP = YP = ZP = 0.5\,\text{m}$.

   Fields of the two positive charges cancel

   The two $+\sqrt{2}\,\text{C}$ charges lie at opposite ends of the hypotenuse, so $P$ is exactly between them and on the line joining them. Each produces a field of equal magnitude directed away from itself; being back-to-back along the same line, they are equal and opposite at $P$ and cancel:

   $E_{\sqrt{2}} = \frac{kq}{r^2} = \frac{9\times10^{9}\times\sqrt{2}}{(0.5)^2}\;\text{each, but they sum to zero.}$ …

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4. KCET 2025Set D-41 markMCQ
   Q.Which of the following statements is not true? (A) Work done to move a charge on an equipotential surface is not zero (B) Equipotential surfaces are the surfaces where the potential is constant (C) Equipotential surfaces for a uniform electric field are parallel and equidistant from each other (D) Electric field is always perpendicular to an equipotential surfaces.

   ›Reveal solutionSolution

   Work on an equipotential surface is $W = q\,\Delta V = 0$ by definition, so the statement that it is 'not zero' is the false one.

   Step 1 — Definition of an equipotential surface.

   An equipotential surface is a surface on which the electrostatic potential $V$ has the same value at every point. This makes statement (B) true — it is simply the definition.

   Step 2 — Test statement (A): work done moving a charge on such a surface.

   The work done by an external agent in moving a charge $q$ from point $A$ to point $B$ is

   $W = q\,(V_B - V_A) = q\,\Delta V$

   If $A$ and $B$ both lie on the same equipotential surface, then $V_A = V_B$, so

   $\Delta V = 0 \quad\Longrightarrow\quad W = q \times 0 = \boxed{0}$

   The work is zero. Statement (A) asserts it is not zero — therefore (A) is FALSE, and since the question asks which statement is not true, (A) is the answer.

   Step 3 — Confirm (D) is true (field ⊥ surface).

   If $\vec{E}$ had any component $E_\parallel$ along the surface, moving a charge a distance $d\ell$ along the surface would do work $W = qE_\parallel\,d\ell \neq 0$, which would mean a potential difference between two points of the surface — contradicting its being equipotential. Hence $E_\parallel = 0$, i.e. $\vec{E}$ is always normal (perpendicular) to the equipotential surface. (D) is true.

   Step 4 — Confirm (C) is true (uniform field case). …

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5. COMEDK 2025Set 2025-M1 markMCQ
   Q.Two identical conducting balls having positive charges $q_1$ and $q_2$ are separated by a distance $r$. If they are made to touch each other and then separated to the same distance, the force between them will be (A) Same as before (B) zero (C) Less than before (D) More than before

   ›Reveal solutionSolution

   When two identical charged conductors touch, they share charge equally, and the electrostatic force after separation is always greater than or equal to the original force unless one charge was zero. The correct option is (D).

   The key idea here is charge redistribution upon contact. When two identical conducting spheres touch, they become a single conductor, and charge flows until both have the same potential. Since they are identical, this means they end up with equal charges. The force between them after separation depends on the product of the new charges compared to the original product.

   ---

   1. Original force The initial electrostatic force between the balls is given by Coulomb’s law:

   $F_{\text{initial}} = k \frac{q_1 q_2}{r^2}$

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

   2. Charge redistribution on contact When the two identical conducting balls touch, the total charge $q_1 + q_2$ is shared equally because they are identical in size and shape. So each ball ends up with:

   $q' = \frac{q_1 + q_2}{2}$

   3. Force after separation After they are separated back to the same distance $r$, the new force is:

   $F_{\text{new}} = k \frac{(q')^2}{r^2} = k \frac{\left(\frac{q_1+q_2}{2}\right)^2}{r^2} = k \frac{(q_1+q_2)^2}{4r^2}$

   4. Compare $F_{\text{new}}$ and $F_{\text{initial}}$ We compare the products:

   $\text{Initial product: } P_i = q_1 q_2$

   $\text{New product: } P_n = \frac{(q_1+q_2)^2}{4}$

   The difference is:

   $P_n - P_i = \frac{(q_1+q_2)^2}{4} - q_1 q_2 = \frac{q_1^2 + 2q_1q_2 + q_2^2 - 4q_1q_2}{4} = \frac{q_1^2 - 2q_1q_2 + q_2^2}{4} = \frac{(q_1 - q_2)^2}{4}$

   Since $(q_1 - q_2)^2 \ge 0$, we have $P_n \ge P_i$. Equality only if $q_1 = q_2$.

   5. Conclusion for the given problem …

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6. KCET 2024Set D-21 markMCQ
   Q.The total electric flux through a closed spherical surface of radius ‘r’ enclosing an electric dipole of dipole moment $2aq$ is (Given $\epsilon_0 =$ permittivity of free space) (A) Zero (B) $\dfrac{q}{\epsilon_0}$ (C) $\dfrac{2q}{\epsilon_0}$ (D) $\dfrac{8\pi r^2 q}{\epsilon_0}$

   ›Reveal solutionSolution

   Gauss's law counts only the net enclosed charge, and a dipole's net charge is zero.

   Step 1 — State Gauss's law

   For any closed surface $S$,

   $\oint_S \vec{E}\cdot d\vec{A} = \frac{q_{\text{enc}}}{\epsilon_0}$

   The crucial point of this law is that the right-hand side depends only on the algebraic sum of the charges inside — not on their positions, not on the size or shape of the surface, and not on any charges outside.

   Step 2 — Find the charge enclosed by the sphere

   A dipole of moment $p = 2aq$ consists of two point charges $+q$ and $-q$ separated by $2a$. Since the whole dipole lies inside the spherical surface, both charges are enclosed:

   $q_{\text{enc}} = (+q) + (-q) = 0$

   Step 3 — Apply Gauss's law

   $\phi = \frac{q_{\text{enc}}}{\epsilon_0} = \frac{0}{\epsilon_0} = 0$

   Step 4 — Why this is not a trick …

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7. KCET 2024Set D-21 markMCQ
   Q.Under electrostatic condition of a charged conductor, which among the following statements is true? (A) The electric field on the surface of a charged conductor is $\dfrac{\sigma}{2\epsilon_0}$, where $\sigma$ is the surface charge density (B) The electric potential inside a charged conductor is always zero (C) Any excess charge resides on the surface of the conductor (D) The net electric field is tangential to the surface of the conductor

   ›Reveal solutionSolution

   Under electrostatic conditions, a conductor’s excess charge sits only on its surface, making option (C) the correct statement.

   The key idea is that in electrostatics, charges inside a conductor are free to move until they reach equilibrium. Once equilibrium is reached, the electric field inside the conductor must be zero everywhere. If there were any field, charges would keep moving. This single fact — zero internal field — forces all the other properties: the surface is an equipotential, the field just outside is perpendicular to it, and any extra charge cannot stay inside.

   Let’s examine each statement carefully.

   1. Option (A): The electric field on the surface of a charged conductor is $\dfrac{\sigma}{2\epsilon_0}$.

      This formula is actually the field due to an infinite sheet of charge. For a conductor, the field just outside the surface is $\dfrac{\sigma}{\epsilon_0}$, not half that. Why? Because the field inside the conductor is zero, so all the flux through a Gaussian pillbox that straddles the surface comes from the outside face. That gives $\dfrac{\sigma}{\epsilon_0}$. The $\dfrac{\sigma}{2\epsilon_0}$ result applies to a thin sheet of charge in free space, where field exists on both sides. Here, the conductor’s interior has no field, so the outside field is double that. So (A) is false.

   2. Option (B): The electric potential inside a charged conductor is always zero.

      Potential is defined relative to a reference (usually infinity). Inside a conductor in equilibrium, the field is zero, so the potential is constant throughout — but that constant need not be zero. It could be positive or negative depending on the charge and the reference. For example, a positively charged isolated conductor has a positive potential. So (B) is false.

   3. Option (C): Any excess charge resides on the surface of the conductor. …

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8. KCET 2024Set D-21 markMCQ
   Q.A cube of side 1 cm contains 100 molecules each having an induced dipole moment of $0.2 \times 10^{-6}\,\text{C} - \text{m}$ in an external electric field of $4\,\text{NC}^{-1}$. The electric susceptibility of the material is _____ $\text{C}^2\text{N}^{-1}\text{m}^{-2}$. (A) 50 (B) 5 (C) 0.5 (D) 0.05

   ›Reveal solutionSolution

   Get the polarisation (total dipole moment per unit volume), then divide by the applied field, since the stated unit $\text{C}^2\text{N}^{-1}\text{m}^{-2}$ means $\chi$ is defined here by $P=\chi E$.

   Step 1 — The concept: polarisation

   When a dielectric is placed in an external field, each molecule acquires an induced dipole moment. The macroscopic measure of this is the polarisation vector

   $P = \frac{\text{net dipole moment}}{\text{volume}}$

   and for a linear dielectric $P$ is proportional to the field, the constant of proportionality being the electric susceptibility.

   Step 2 — Volume of the cube

   $a = 1\ \text{cm} = 10^{-2}\ \text{m} \quad\Rightarrow\quad V = a^3 = \left(10^{-2}\right)^3 = 10^{-6}\ \text{m}^3$

   Step 3 — Total dipole moment

   The $100$ molecules each carry $p = 0.2\times10^{-6}\ \text{C\,m}$ and (being induced by the same field) are aligned, so their moments simply add:

   $p_{\text{total}} = N p = 100 \times 0.2\times10^{-6} = 2\times10^{-5}\ \text{C\,m}$

   Step 4 — Polarisation

   $P = \frac{p_{\text{total}}}{V} = \frac{2\times10^{-5}}{10^{-6}} = 20\ \text{C\,m}^{-2}$

   Step 5 — Susceptibility …

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9. COMEDK 2024Set 2024-A1 markMCQ
   Q.3 point charges each of $-\mathrm{q}$ are placed on the circumference of a circle of diameter $2 \mathrm{a}$ at $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ respectively as shown in figure. The electric field at $\mathrm{O}$ is (A) Zero (B) $\frac{\mathrm{Kq}}{2 \mathrm{a}}$ towards $\mathrm{OC}$ (C) $\frac{\mathrm{Kq}}{\mathrm{a}}$ towards $\mathrm{OB}$ (D) $\frac{\mathrm{Kq}}{\mathrm{a}}$ towards $\mathrm{OA}$

   ›Reveal solutionSolution

   Two of the three equal charges sit at opposite ends of a diameter and their fields at O cancel; only the third charge (at B) is left, so the net field points towards OB - option (C).

   The field at O is the vector sum of the fields of the three charges. Each charge is $-q$ at a distance equal to the radius $a$ (diameter $=2a$), and for a negative charge the field at O points toward that charge.

   Charges A and C lie at opposite ends of a diameter, so their fields at O are equal in magnitude and exactly opposite in direction:

   $\vec{E}_A+\vec{E}_C=\frac{Kq}{a^2}\,\hat{u}_{OA}+\frac{Kq}{a^2}\,\hat{u}_{OC}=\vec{0}.$

   Charge B is then unopposed, so the resultant is its field alone, directed from O toward B:

   $\vec{E}_O=\frac{Kq}{a^2}\ \text{along } OB.$ …

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10. COMEDK 2024Set 2024-E1 markMCQ
    Q.Two point charges $\mathrm{M}$ and $\mathrm{N}$ having charges $+q$ and $-q$ respectively are placed at a distance apart. Force acting between them is $\mathrm{F}$. If $30 \%$ of charge of $\mathrm{N}$ is transferred to $\mathrm{M}$, then the force between the charges becomes: (A) F (B) $\frac{100}{49} F$ (C) $\frac{49}{100} F$ (D) $\frac{9}{16} F$

    ›Reveal solutionSolution

    Transferring $30\%$ of $N$'s charge leaves both magnitudes at $0.7q$, so the force scales by $0.7\times0.7=0.49$, giving $\dfrac{49}{100}F$.

    Initial force between $+q$ and $-q$ separated by $r$:

    $F=\frac{1}{4\pi\varepsilon_0}\frac{q\cdot q}{r^{2}}.$

    $30\%$ of $N$'s charge is $0.3q$, transferred from $N$ to $M$:

    $q_M=q-0.3q=0.7q,\qquad q_N=q-0.3q=0.7q\ \text{(magnitude).}$ …

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11. COMEDK 2024Set 2024-E1 markMCQ
    Q.Three point charges are located on a circular arc at $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ as shown in the figure below. The total electric field at the centre of the $\operatorname{arc}(\mathrm{C})$ is (A) $15000 \mathrm{~NC}^{-1}$ (B) $10000 \mathrm{~N} \mathrm{C}^{-1}$ (C) $20000 \mathrm{~NC}^{-1}$ (D) $5000 \mathrm{~NC}^{-1}$

    ›Reveal solutionSolution

    The two equal positive charges give $10000$ N/C net along the axis and the negative charge $5000$ N/C the other way; resultant $=5000$ N/C.

    With $r=6.0$ cm $=0.06$ m, $k=9\times10^9$:

    $E_{+4}=\frac{(9\times10^9)(4\times10^{-9})}{(0.06)^2}=\frac{36}{0.0036}=10000\text{ N/C}$

    $E_{-2}=\frac{(9\times10^9)(2\times10^{-9})}{(0.06)^2}=5000\text{ N/C}$ …

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12. COMEDK 2024Set 2024-M1 markMCQ
    Q.Five charges, '$q$' each are placed at the comers of a regular pentagon of side '$a$' as shown in figure. First, charge from '$A$' is removed with other charges intact, then charge at '$A$' is replaced with an equal opposite charge. The ratio of magnitudes of electric fields at $\mathrm{O}$, without charge at $A$ and that with equal and opposite charge at $A$ is (A) 4 : 1 (B) 2 : 1 (C) 1 : 4 (D) 1 : 2

    ›Reveal solutionSolution

    The key idea is that the electric field at the centre of a regular pentagon due to five equal charges is zero by symmetry. Removing one charge leaves a net field equal in magnitude to the field of that missing charge alone. Replacing it with an equal opposite charge doubles that field. The ratio of magnitudes is therefore 1 : 2, corresponding to option (D).

    Concept and intuition:

    For a regular pentagon, the centre is equidistant from all vertices. If five equal charges are placed at the vertices, the vector sum of the electric fields at the centre is zero — each field is cancelled by the others due to symmetry. This is the same principle that makes the field at the centre of a regular polygon with equal charges zero.

    Now, if we remove one charge, the symmetry is broken. The net field at O is exactly the field that the missing charge would have produced (since the remaining four still sum to the negative of that one field). If we instead replace the removed charge with an equal opposite charge, that charge now contributes a field in the same direction as the field from the missing charge (because the sign flips), so the total field becomes twice the field of a single charge. The ratio of the magnitudes (first case : second case) is therefore 1 : 2.

    ---

    Step-by-step reasoning:

    1. Original configuration (five equal charges): The centre O is equidistant from all vertices. Let this distance be $r$ (the circumradius). Each charge $q$ produces a field of magnitude $E_0 = \frac{kq}{r^2}$ at O, directed radially outward (if $q>0$). Because the pentagon is regular, the five vectors are equally spaced in direction by $72^\circ$. Their vector sum is zero:

    $\vec{E}_{\text{total}} = \vec{0}.$

    2. First scenario: charge at A is removed. Now only four charges remain (at B, C, D, E). Their net field at O is no longer zero. However, since the original five fields summed to zero, the sum of the four remaining fields must be exactly the negative of the field that was produced by the charge at A:

    $\vec{E}_{\text{four}} = -\vec{E}_A.$

    The magnitude is therefore $|\vec{E}_{\text{four}}| = E_0$.

    3. Second scenario: charge at A is replaced with an equal opposite charge $-q$. …

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