Q.Two point charges $+15\,\mu\text{C}$ and $-10\,\mu\text{C}$ are separated by a distance of 20 cm in air. Calculate the electric field at the mid point of line joining two charges. If a point charge of 20 mC is placed at that mid-point. What is the magnitude of electric force experienced by it?

Showing the 12 most recent of 15 on this concept.

1. CBSE 2025Set D1 markMCQ
   Q.Which of the following has unit volt-metre^-1? (A) Electric flux (B) Electric potential (C) Electric field (D) Electric capacity

   ›Reveal solutionSolution

   Volt per metre (V/m) is the unit of electric field.

   Electric field is force per unit charge, E = F/q, giving N/C. It is also the negative gradient of potential, E = −dV/dx, giving volt/metre. These two units are identical: 1 N/C = 1 V/m.

   …

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2. CBSE 2025Set ANNUAL1 markMCQ
   Q.An electron placed in an electric field experiences an electrical force equal to its weight. The magnitude of the field is

   (a) mge

   (b) mg/e

   (c) e/mg

   (d) e^2g/2m

   ›Reveal solutionSolution

   When the electric force on a charge exactly balances its weight, eE = mg, so the field magnitude is E = mg/e.

   An electron of mass m and charge magnitude e placed in a uniform electric field E experiences an electric force of magnitude F = eE.

   It is given that this electric force equals the electron's weight:

   eE = mg

   Solving for E:

   E = mg/e

   …

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3. CBSE 2025Set ANNUAL1 markMCQ
   Q.The electrostatic force per unit charge is known as ______.

   (a) Electric current

   (b) Electric potential

   (c) Electric field

   ›Reveal solutionSolution

   The electrostatic force experienced per unit positive test charge, placed at a point, is by definition the electric field at that point.

   The electric field $\vec{E}$ at a point is defined operationally as the force $\vec{F}$ that a small positive test charge $q_0$ would experience at that point, divided by the magnitude of the test charge:

   $\vec{E} = \dfrac{\vec{F}}{q_0}$ …

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4. CBSE 2025Set ANNUAL1 mark
   Q.The test charge used to measure electric field at a point should be vanishingly small. Why?

   ›Reveal solutionSolution

   A vanishingly small test charge is used so it does not alter the very field/source-charge distribution it is being used to measure.

   Electric field at a point is defined as $\vec{E} = \lim_{q_0 \to 0} \frac{\vec{F}}{q_0}$, where $q_0$ is a small positive test charge placed at that point.

   If the test charge were large, it would exert a significant force of its own on the source charge(s) creating the field, causing them to shift position (if free to move) or altering the charge distribution. This would change the very field being measured, giving an inaccurate value. By taking the limit as $q_0 \to 0$, we ens …

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5. CBSE 2024Set ANNUAL1 markMCQ
   Q.The electrostatic force experienced by a unit positive charge at a point in space is called -

   (a) Electric Current

   (b) Electric Potential

   (c) Electric Field

   (d) Electric Space

   ›Reveal solutionSolution

   The force per unit positive test charge at a point defines the electric field there.

   By definition, the electric field $\vec{E}$ at a point in space is the electrostatic force $\vec{F}$ experienced by a very small unit positive test charge placed at that point: $\vec{E} = \vec{F}/q_0$. This is distinct from …

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6. CBSE 2024Set A1 markMCQ
   Q.The dimensional formula of intensity of electric field is (A) [MLT^-2 A^-1] (B) [MLT^-3 A^-1] (C) [MLT^-3 A] (D) [ML^2 T^-3 A^-1]

   ›Reveal solutionSolution

   E = Force/charge → [MLT⁻²]/[AT] = [MLT⁻³A⁻¹].

   Electric field intensity is force per unit charge: $E = \dfrac{F}{q}$.

   Dimensions of force = [MLT⁻²]. Dimensions of charge = current × time = [AT]. …

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7. CBSE 2024Set ANNUAL1 markMCQ
   Q.Which one of the following is the unit of Electric field?

   (a) Coulomb

   (b) Newton

   (c) Volt

   (d) NC⁻¹

   ›Reveal solutionSolution

   Electric field is force per unit charge, so its SI unit follows directly from N/C.

   Electric field at a point is defined as the force experienced per unit positive test charge placed there:

   $\vec{E} = \dfrac{\vec{F}}{q_0}$

   Force is measured in newtons (N) and charge in coulombs (C), so the SI unit of $E$ is newton per coulomb (NC⁻¹) — this is numerically the same as volt per metre (V/m), but NC⁻¹ is the unit that comes directly from the defining equation.

   …

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8. CBSE 2023Set 55/3/11 markMCQ
   Q.An electron experiences a force $(1.6\times10^{-16}\ \text{N})\hat{i}$ in an electric field $\vec{E}$. The electric field $\vec{E}$ is :

   (a) $\left(1.0\times10^{3}\ \dfrac{\text{N}}{\text{C}}\right)\hat{i}$

   (b) $-\left(1.0\times10^{3}\ \dfrac{\text{N}}{\text{C}}\right)\hat{i}$

   (c) $\left(1.0\times10^{-3}\ \dfrac{\text{N}}{\text{C}}\right)\hat{i}$

   (d) $-\left(1.0\times10^{-3}\ \dfrac{\text{N}}{\text{C}}\right)\hat{i}$

   ›Reveal solutionSolution

   The force on a charge in an electric field is $\vec{F} = q\vec{E}$; since the electron has negative charge and the force points in $+\hat{i}$, the field must point in $-\hat{i}$ with magnitude $1.0 \times 10^3 \, \text{N/C}$. The answer is (b).

   The relationship between electric force and electric field is one of the most fundamental in electrostatics. A charge $q$ placed in an electric field $\vec{E}$ experiences a force given by

   $\vec{F} = q\vec{E}$

   This is the defining equation for the electric field: the field is the force per unit positive charge. The key insight here is that the sign of the charge matters. A positive charge experiences force in the same direction as the field, but a negative charge experiences force opposite to the field direction.

   An electron carries charge $q = -e = -1.6 \times 10^{-19} \, \text{C}$.

   Let me work through this systematically:

   1. Write down what we know.

      The force on the electron is $\vec{F} = (1.6 \times 10^{-16} \, \text{N})\hat{i}$, pointing in the positive $x$-direction. The electron's charge is $q = -1.6 \times 10^{-19} \, \text{C}$.

   2. Apply the force-field relationship.

      From $\vec{F} = q\vec{E}$, we can solve for the electric field:

   $\vec{E} = \frac{\vec{F}}{q}$

   3. Substitute the values.

   $\vec{E} = \frac{(1.6 \times 10^{-16} \, \text{N})\hat{i}}{-1.6 \times 10^{-19} \, \text{C}}$

   4. Simplify the magnitude. …

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9. CBSE 2023Set F1 markMCQ
   Q.The force, acting on per unit charge is called (A) Electric current (B) Electric potential (C) Electric field (D) Electric space

   ›Reveal solutionSolution

   Electric field E = F/q₀ — the force experienced per unit positive test charge.

   The electric field at a point is defined as the electrostatic force experienced by a small positive test charge placed there, divided by the magnitude of that charge:

   $\vec{E} = \dfrac{\vec{F}}{q_0}$

   …

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10. CBSE 2023Set F1 markMCQ
    Q.Which of the following physical quantities is a vector? (A) Electric flux (B) Electric potential (C) Electric potential energy (D) Electric intensity

    ›Reveal solutionSolution

    Electric intensity (field) is a vector; flux, potential, PE are scalars.

    • Electric flux $\Phi = \vec E\cdot\vec A$ is a dot product → scalar.
    • Electric potential $V$ → scalar (energy per unit charge).
    • Electric potential energy → scalar (energy). …

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11. CBSE 2022Set I1 markMCQ
    Q.Intensity of electric field at a point is (A) E = Fq (B) E = F/q (C) E = ½Fq (D) E = q/F

    ›Reveal solutionSolution

    Electric field intensity E = F/q.

    The electric field at a point is defined as the force experienced per unit positive test charge placed at that point:

    $\vec{E} = \frac{\vec{F}}{q}$

    …

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12. CBSE 2022Set I1 markMCQ
    Q.The S.I. unit of electric intensity is (A) NC (B) N/C (C) N.C^2 (D) N/C^2

    ›Reveal solutionSolution

    Electric intensity (field) E = F/q, so its SI unit is N/C.

    Electric field intensity is defined as the force per unit positive test charge:

    $E = \dfrac{F}{q}$

    …

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