Q.Two spheres carrying charges +6 μC and +9 μC, separated by a distance d, experience a force of repulsion F. When a charge of −3 μC is added to each sphere and distance d is kept the same, the new force of repulsion will be

Showing the 12 most recent of 40 on this concept.

1. CBSE 2026Set A1 markMCQ
   Q.Coulomb's law is valid for (A) Point charges only (B) Dispersed charges only (C) Both point charges and dispersed charges (D) Neutral particles

   ›Reveal solutionSolution

   Coulomb's law is defined for point charges; extended bodies need integration.

   Coulomb's law states $F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r^2}$, where $r$ is the distance between the charges.

   …

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2. CBSE 2026Set ANNUAL1 markMCQ
   Q.Two spheres carrying charges +6 μC and +9 μC, separated by a distance d, experience a force of repulsion F. When a charge of −3 μC is added to each sphere and distance d is kept the same, the new force of repulsion will be

   (a) 3F

   (b) F/9

   (c) F

   (d) F/3

   ›Reveal solutionSolution

   New force = F/3, because force is proportional to the product of the charges and only the charges changed, not the distance.

   By Coulomb's law, the force between two point charges $q_1$ and $q_2$ separated by a fixed distance $d$ is

   $F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{d^2}$

   Originally $q_1 = +6\ \mu C$ and $q_2 = +9\ \mu C$, so $F \propto q_1 q_2 = 54$ (in $\mu C^2$).

   After $-3\ \mu C$ is added to each sphere: …

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3. CBSE 2026Set ANNUAL1 markMCQ
   Q.Two sphere of charge 2μc and 3μc are located at a distance 20 cm apart in air. The ratio of magnitude of electric forces acting between these spheres will be

   (a) 1 : 1

   (b) 2 : 3

   (c) 3 : 2

   (d) 4 : 9

   ›Reveal solutionSolution

   The mutual electric force between two charges is an action-reaction pair, so both spheres feel equal magnitude forces regardless of the charge values.

   By Coulomb's law the force sphere 1 exerts on sphere 2 has magnitude F = k q1 q2 / r^2, and the force sphere 2 exerts on sphere 1 has the same magnitude k q1 q2 / r^2, just opposite in direction (Newton's third law applies to electrostatic forces just as it do …

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4. CBSE 2025Set X11 markMCQ
   Q.A point charge $q_1$ exerts a force $F$ on another point charge $q_2$ when placed at a fixed distance. If another point charge $q_3$ is brought near $q_2$, the force on $q_2$ due to $q_1$ :

   (a) increases

   (b) decreases

   (c) may increase or decrease

   (d) does not change

   ›Reveal solutionSolution

   (d) does not change. By the principle of superposition, the electrostatic force between $q_1$ and $q_2$ is given by Coulomb's law $F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r^2}$ and depends …

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5. CBSE 2025Set D1 markMCQ
   Q.The distance between two charges is made half and one of the charges is also halved. The force acting between the two will become as compared to previous value (A) half (B) double (C) thrice (D) none of these

   ›Reveal solutionSolution

   Coulomb force F ∝ q₁q₂/r²; halving one charge (×½) and halving the distance (×4) gives a net factor of 2, so the force doubles.

   Coulomb's law:

   $F = \frac{k q_1 q_2}{r^2}$

   Initial force: $F = \dfrac{k q_1 q_2}{r^2}$.

   Now one charge becomes $q_1/2$ and the distance becomes $r/2$:

   …

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6. CBSE 2025Set D1 markMCQ
   Q.On inserting a dielectric material between two positive charges in air, the value of repulsive force will (A) increase (B) decrease (C) remain same (D) become zero

   ›Reveal solutionSolution

   A dielectric weakens the field between charges by a factor K, so the repulsive force falls to F₀/K.

   The Coulomb force between two charges in air is F₀ = (1/4πε₀)·q₁q₂/r². Filling the space with a dielectric of relative permittivity K replaces ε₀ with Kε₀:

   F = (1/4πKε₀)·q₁q₂/r² = F₀/K

   …

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7. CBSE 2025Set ANNUAL1 markMCQ
   Q.The law governing the force between static electric charges is known as

   (i) Ampere's law

   (ii) Ohm's law

   (iii) Faraday's law

   (iv) Coulomb's law

   ›Reveal solutionSolution

   The force between two static (point) electric charges is governed by Coulomb's law.

   Ampere's law relates a magnetic field to the current producing it, Ohm's law relates current and voltage in a conductor, and Faraday's law deals with electromagnetic induction. None of these describes the force between charges at rest.

   …

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8. CBSE 2024Set IMPROVEMENT1 markMCQ
   Q.On placing dielectric material between two point charges in air, repulsive force between them will —

   (a) Increase

   (b) Decrease

   (c) Remain same

   (d) Zero

   ›Reveal solutionSolution

   Placing a dielectric between two charges reduces the force between them.

   …

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9. CBSE 2024Set FS1 markMCQ
   Q.Force of 80 Newton works between two point charges placed at a fixed distance apart in air. When these charges are placed at the same distance apart in a dielectric medium, then force of 8 Newton works on it. The dielectric constant of medium will be:

   (i) $K = -10$

   (ii) $K = 10$

   (iii) $K = 0.01$

   (iv) $K = -0.01$

   ›Reveal solutionSolution

   $K=\dfrac{F_{air}}{F_{medium}}=\dfrac{80}{8}=10$ — option (ii).

   Concept. Coulomb's force between two charges at separation $r$ is

   $F_{air}=\frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r^2},\qquad F_{medium}=\frac{1}{4\pi\varepsilon_0 K}\frac{q_1q_2}{r^2}.$

   Placing a dielectric of constant $K$ reduces the force by the factor $K$. …

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10. CBSE 2024Set A1 mark
    Q.Match Column 'A' with Column 'B' and write the correct pair. Column 'A' item: 'Electrostatic force'. Column 'B' options:

    (i) De-Broglie

    (ii) Maxwell

    (iii) Ohm

    (iv) Einstein

    (v) Coulomb

    (vi) Lenz

    (vii) Young.

    ›Reveal solutionSolution

    Electrostatic force is governed by Coulomb's law.

    The force of attraction or repulsion between two stationary point charges is called the electrostatic (or Coulomb) force, and its magnitude is given by Coulomb's law:

    $F = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r^2}$

    …

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11. CBSE 2024Set ANNUAL1 markMCQ
    Q.Two charged spheres are separated by a distance d, exert a force F on each other. If the charges are doubled and the distance between them is doubled then the force is

    (a) F

    (b) F/2

    (c) F/4

    (d) 4F

    ›Reveal solutionSolution

    Coulomb's law force scales as (charge product)/(distance)^2; doubling both charges and the distance leaves the force unchanged.

    By Coulomb's law, the force between two point charges q1 and q2 separated by distance d is

    $F = \dfrac{1}{4\pi\epsilon_0}\dfrac{q_1 q_2}{d^2} = k\dfrac{q_1q_2}{d^2}$

    …

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12. CBSE 2024Set ANNUAL1 mark
    Q.What is the name of the electrical force acting between two charges at rest?

    ›Reveal solutionSolution

    The force between two charges at rest is the electrostatic (Coulomb) force.

    The electrical force acting between two charges that are at rest (not moving) is called the electrostatic force, or Coulomb force, since it is governed by Coulomb's law:

    $F = \dfrac{1}{4\pi\epsilon_0}\dfrac{q_1 q_2}{r^2}$

    …

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