Q.Define electric potential at a point in an electrostatic field, and state its SI unit.

Showing the 12 most recent of 44 on this concept.

1. CBSE 2026Set 55/1/11 markMCQ
   Q.A conducting wire connects two charged metallic spheres A and B of radii $r_1$ and $r_2$ respectively. The distance between the spheres is very large compared to their radii. The ratio of electric fields, $\left(\dfrac{E_A}{E_B}\right)$ at the surfaces of spheres A and B will be (A) $\dfrac{r_1}{r_2}$ (B) $\dfrac{r_2}{r_1}$ (C) $\dfrac{r_1^2}{r_2^2}$ (D) $\dfrac{r_2^2}{r_1^2}$

   ›Reveal solutionSolution

   When two widely separated conducting spheres are connected by a wire, they reach the same electric potential. Since surface field $E = \frac{kQ}{r^2}$ and potential $V = \frac{kQ}{r}$, combining these gives $E \propto 1/r$. Therefore the ratio of surface fields is $E_A/E_B = r_2/r_1$, which corresponds to option (B).

   The key insight here is about what happens when conductors are connected by a wire. Charge flows until both spheres are at the same electric potential — that's the fundamental condition for electrostatic equilibrium in a conductor. Once you grasp that, the rest is just algebra.

   Let's think about why potential equality is the right starting point. A conducting wire means the two spheres form a single conductor. In electrostatics, the entire surface of a conductor is an equipotential. So spheres A and B must have the same potential $V$.

   Now, for an isolated conducting sphere of radius $r$ carrying charge $Q$, the potential at its surface (taking infinity as zero) is:

   $V = \frac{1}{4\pi\epsilon_0} \frac{Q}{r}$

   And the electric field just outside its surface is:

   $E = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2}$

   Notice the relationship: $E = V/r$. That's a neat shortcut we'll use.

   Tip

   For any isolated conducting sphere, $E = V/r$ directly. This saves you from carrying the $Q$ through the algebra — just remember it comes from $V = kQ/r$ and $E = kQ/r^2$.

   Let's work through it step by step.

   1. Set potentials equal. Since the wire connects them, $V_A = V_B$. Using $V = kQ/r$ (where $k = 1/4\pi\epsilon_0$):

   $k\frac{Q_A}{r_1} = k\frac{Q_B}{r_2}$

   Cancel $k$ and rearrange:

   $\frac{Q_A}{Q_B} = \frac{r_1}{r_2}$

   2. Write the surface field ratio. For each sphere, $E = kQ/r^2$. So:

   $\frac{E_A}{E_B} = \frac{k Q_A / r_1^2}{k Q_B / r_2^2} = \frac{Q_A}{Q_B} \cdot \frac{r_2^2}{r_1^2}$

   3. Substitute the charge ratio. From step 1, $Q_A/Q_B = r_1/r_2$:

   $\frac{E_A}{E_B} = \frac{r_1}{r_2} \cdot \frac{r_2^2}{r_1^2} = \frac{r_2}{r_1}$

   That's it. The larger sphere has the smaller surface field.

   Watch out

   A common mistake is to assume the charges become equal (they don't — the larger sphere holds more charge) or to directly use $E \propto 1/r^2$ without accounting for the charge redistribution. Always start from potential equality, not charge equality.

   Note

   The condition "distance between spheres is very large compared to their radii" ensures we can treat each sphere as isolated — no mutual induction effects. If they were close, the charge distribution would become non-uniform and this simple analysis would break down.

   ✓Final answer

   The correct option is (B) $\dfrac{r_2}{r_1}$.

2. CBSE 2026Set 55/3/11 markMCQ
   Q.A particle of mass $m$ and charge $q$ starts from rest and moves in an electric field $\vec{E} = E_0\,\hat{i}$. After travelling a distance $x$ in the field along the $x$-axis, the kinetic energy of the particle will be : (A) $q\,E_0\,x^2$ (B) $q\,E_0\,x$ (C) $q^2\,E_0\,x$ (D) $2\,q^2\,E_0\,x$

   ›Reveal solutionSolution

   Work done by a constant electric field equals force times displacement; since the particle starts from rest, all that work converts to kinetic energy, giving $K = q E_0 x$.

   The heart of this problem is the work-energy theorem: the work done by all forces on a particle equals its change in kinetic energy. When a charged particle moves through an electric field, the field exerts a force that does work, and if the particle starts from rest, every joule of work becomes kinetic energy.

   A uniform electric field $\vec{E} = E_0\,\hat{i}$ exerts a force $\vec{F} = q\vec{E}$ on a charge $q$. This force is constant in magnitude and direction, so the work done is simply force times displacement along the direction of the force.

   Step-by-step reasoning

   1. Identify the force on the particle. The electric force on a charge $q$ in field $\vec{E}$ is

   $\vec{F} = q\vec{E} = q E_0\,\hat{i}$

   The magnitude is $F = q E_0$, directed along the positive $x$-axis.

   2. Calculate the work done by this force. The particle moves a distance $x$ along the $x$-axis, in the same direction as the force. Work done by a constant force is

   $W = F \cdot d = q E_0 \cdot x = q E_0 x$

   3. Apply the work-energy theorem. The particle starts from rest, so initial kinetic energy $K_i = 0$. The work-energy theorem states

   $W = \Delta K = K_f - K_i$

   Therefore,

   $K_f = W = q E_0 x$

   The kinetic energy after travelling distance $x$ is simply the work done by the electric field.

   Watch out

   A common mistake is to confuse the distance $x$ with the square of distance. The work done by a constant force is linear in displacement, not quadratic. Option (A) incorrectly includes $x^2$, which would arise only if the force itself depended on position — but here $E_0$ is constant.

   Tip

   You can also verify dimensions: kinetic energy has units of energy (joules). Check option (B): $[q E_0 x] = \text{C} \cdot (\text{N/C}) \cdot \text{m} = \text{N·m} = \text{J}$ ✓. Option (A) would give $\text{J·m}$, which is wrong.

   ✓Final answer

   The kinetic energy of the particle after travelling distance $x$ is $\boxed{q E_0 x}$, so the correct option is (B).

3. CBSE 2026Set ANNUAL1 markMCQ
   Q.SI unit of electric potential is:

   (a) Ohm

   (b) Volt

   (c) Coulomb

   (d) Ampere

   ›Reveal solutionSolution

   Electric potential is defined as work done per unit charge, so its SI unit is the Volt.

   Electric potential at a point is $V = \dfrac{W}{q}$, i.e. the work done in bringing a unit positive charge from infinity to that point. Since work is measured in joules (J) and charge in coulombs (C), the unit of potential is J/C, which is given the special name Volt (V).

   ✓Final answer

   SI unit of electric potential is the Volt (option b).

4. CBSE 2026Set ANNUAL1 markMCQ
   Q.[Case study, continued] The charges q1 and q2 produce a potential, which at any point P will be

   (a) V1,2 = 1/(4πε₀) × (q1/r1P² + q2/r2P²)

   (b) V1,2 = 1/(4πε₀) × (q1/r1P + q2/r2P)

   (c) V1,2 = 1/(4πε₀) × (q1/r1P - q2/r2P)

   (d) V1,2 = 1/(4πε₀) × (2q1/r1P + 3q2/r2P)

   ›Reveal solutionSolution

   The potential due to a group of point charges at any point is just the plain algebraic (scalar) sum of the potentials each charge produces there — the superposition principle for potential.

   Electric potential obeys the superposition principle: the total potential at any point due to several charges equals the SCALAR sum (not vector sum, since potential is a scalar) of the potentials due to each individual charge, each given by $V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}$ (distance to the first power).

   For two charges $q_1$ (at distance $r_{1P}$ from P) and $q_2$ (at distance $r_{2P}$ from P):

   $V_{1,2} = \frac{1}{4\pi\varepsilon_0}\left(\frac{q_1}{r_{1P}} + \frac{q_2}{r_{2P}}\right)$

   This rules out option (a), which wrongly squares the distances (that would be a field-like expression, not potential), and option (d), which has arbitrary extra numerical coefficients (2, 3) not present in the actual physics. Option (c) wrongly subtracts instead of adding (there is no reason to subtract unless one charge is explicitly negative, which is already accounted for by the sign of $q_2$ itself).

   ✓Final answer

   (b) $V_{1,2} = \dfrac{1}{4\pi\varepsilon_0}\left(\dfrac{q_1}{r_{1P}} + \dfrac{q_2}{r_{2P}}\right)$

5. CBSE 2026Set ANNUAL1 markMCQ
   Q.The standard potential of earth is :

   (a) Zero

   (b) Infinite

   (c) One

   (d) None of the above

   ›Reveal solutionSolution

   The Earth is the chosen reference for potential, so its standard potential is taken as zero.

   Electric potential is always measured relative to some reference. Because the Earth is a very large conductor whose potential is practically unaffected by adding or removing charge, it is universally chosen as the reference (zero) level of potential.

   Hence the standard potential of the Earth is taken to be zero, and any conductor connected to earth (grounded) is brought to zero potential.

   ✓Final answer

   (a) Zero.

6. CBSE 2026Set ANNUAL1 mark
   Q.Match the Column A with Column B and write the correct pair. Column A: Potential gradient. Column B:

   (i) μ₀nI,

   (ii) Volt × second,

   (iii) μ₀nI/2,

   (iv) Volt × meter,

   (v) Volt × meter⁻¹,

   (vi) Volt.

   ›Reveal solutionSolution

   Potential gradient dV/dx has unit V/m = Volt × metre⁻¹, option (v).

   The potential gradient is the rate of change of electric potential with distance, dV/dx. Its SI unit is volt per metre (V/m), i.e. Volt × metre⁻¹. In magnitude it equals the electric field intensity (E = −dV/dx). This matches Column B entry (v).

   ✓Final answer

   (v) Volt × meter⁻¹.

7. CBSE 2025Set 55/5/11 markMCQ
   Q.The electric field at a point in a region is given by $\vec{E}=\dfrac{\alpha}{r^{2}}\,\hat{r}$ (a radial field), where $\alpha$ is a constant and $r$ is the distance of the point from the origin. The magnitude of the potential at the point is: (A) $\dfrac{\alpha}{r}$ (B) $\dfrac{\alpha r^{2}}{2}$ (C) $\dfrac{\alpha}{2r^{2}}$ (D) $-\dfrac{\alpha}{r}$

   ›Reveal solutionSolution

   For a radial electric field $\vec{E} = \frac{\alpha}{r^2}\,\hat{r}$, integrate $-\vec{E} \cdot d\vec{l}$ along a radial path from infinity to find the potential; the magnitude is $\frac{\alpha}{r}$.

   The connection between electric field and potential is one of the most fundamental relationships in electrostatics. The electric field points in the direction of steepest decrease of potential, and its magnitude tells us how rapidly the potential drops. Mathematically, $\vec{E} = -\nabla V$, or in one dimension, $E = -\frac{dV}{dr}$ for a radial field.

   To find the potential at a point, we integrate the electric field along a path. The potential difference between two points is:

   $V(r) - V(r_0) = -\int_{r_0}^{r} \vec{E} \cdot d\vec{l}$

   We conventionally choose $r_0 = \infty$ as our reference point where $V(\infty) = 0$, so:

   $V(r) = -\int_{\infty}^{r} \vec{E} \cdot d\vec{l}$

   Now let's work through this problem step by step.

   1. Set up the line integral for a radial field.

      Since both $\vec{E}$ and the path element $d\vec{l}$ point radially (we choose a radial path for simplicity), we have:

   $\vec{E} \cdot d\vec{l} = E_r \, dr = \frac{\alpha}{r^2} \, dr$

   2. Evaluate the integral from infinity to $r$.

   $V(r) = -\int_{\infty}^{r} \frac{\alpha}{r^2} \, dr$

   Reversing the limits to make the calculation cleaner:

   $V(r) = \int_{r}^{\infty} \frac{\alpha}{r^2} \, dr$

   3. Perform the integration.

   $V(r) = \alpha \int_{r}^{\infty} \frac{1}{r^2} \, dr = \alpha \left[-\frac{1}{r}\right]_{r}^{\infty}$

   $V(r) = \alpha \left(0 - \left(-\frac{1}{r}\right)\right) = \frac{\alpha}{r}$

   4. Interpret the result.

      The potential is positive when $\alpha > 0$ (a repulsive field, like that of a positive point charge) and negative when $\alpha < 0$. The question asks for the magnitude of the potential, which is $\left|\frac{\alpha}{r}\right|$.

   Watch out

   Don't confuse the sign convention. Option (D) gives $-\frac{\alpha}{r}$, which would be correct only if $\alpha$ itself were defined with the opposite sign convention. The magnitude depends on whether we're asked for $|V(r)|$ or just the functional form.

   Tip

   For any radial field $\vec{E} = f(r)\,\hat{r}$, the potential is always $V(r) = -\int_{\infty}^{r} f(r')\,dr'$. The $1/r^2$ field gives $1/r$ potential, just like a point charge.

   Since the problem asks for "the magnitude of the potential" and our result is $V(r) = \frac{\alpha}{r}$, the magnitude is $\frac{\alpha}{r}$ (assuming $\alpha$ is taken as a positive constant in the context of this problem).

   ✓Final answer

   The correct option is (A) $\dfrac{\alpha}{r}$.

8. CBSE 2025Set IMPROVEMENT1 markMCQ
   Q.Electron volt is the unit of which of the following?

   (a) Electric Potential

   (b) Energy

   (c) Electric Field

   (d) Electric Current

   ›Reveal solutionSolution

   Electron volt (eV) is a unit of energy, not of potential, field or current.

   By definition, 1 electron volt is the amount of kinetic energy gained (or lost) by a single electron when it is accelerated (or decelerated) through a potential difference of 1 volt. Since work done $W = qV$, for $q = e = 1.6\times10^{-19}\,\text{C}$ and $V = 1\,\text{volt}$, $1\ \text{eV} = 1.6\times10^{-19}\,\text{J}$, which is clearly a unit of energy.

   ✓Final answer

   (b) Energy — 1 eV is the kinetic energy gained by an electron accelerated through a potential difference of 1 volt.

9. CBSE 2025Set D1 markMCQ
   Q.The speed of an electron accelerated from rest under a potential difference V is (A) proportional to V (B) proportional to √V (C) proportional to 1/V (D) proportional to V^2

   ›Reveal solutionSolution

   The kinetic energy gained equals eV; since ½mv² = eV, the speed v ∝ √V.

   An electron accelerated from rest through a potential difference V gains kinetic energy equal to the work done by the field:

   $eV = \frac{1}{2}mv^2$

   Solving for the speed:

   $v = \sqrt{\frac{2eV}{m}}$

   Since e and m are constants, $v \propto \sqrt{V}$ — the speed is proportional to the square root of the accelerating voltage.

   ✓Final answer

   (B) proportional to √V.

10. CBSE 2025Set ANNUAL1 markMCQ
    Q.Two charges 1C and -1C are placed 1 m apart. The potential at the centre of the line joining the two charges will be

    (a) 2 V

    (b) -2 V

    (c) zero

    (d) 0.5 V

    ›Reveal solutionSolution

    At the midpoint of the line joining equal-and-opposite charges, the two charges are equidistant, so their potential contributions are equal in magnitude but opposite in sign and cancel exactly.

    Electric potential due to a point charge q at distance r is V = kq/r (k = 1/4-pi-epsilon0).

    The two charges are q1 = +1 C and q2 = -1 C, separated by 1 m, so each is r = 0.5 m from the midpoint.

    V = k(+1)/0.5 + k(-1)/0.5 = 2k - 2k = 0

    Note that potential is a scalar, so it simply adds algebraically (unlike the electric field at that point, which does NOT vanish - the fields from the two charges point in the same direction there and add up).

    ✓Final answer

    (c) zero.

11. CBSE 2025Set ANNUAL1 markMCQ
    Q.When a photon is accelerated (from rest) through a potential difference of one volt, the kinetic energy gained by it is equal to

    (a) 1837eV

    (b) 1 eV

    (c) $\frac{1}{1837}$ eV

    (d) none of the above.

    ›Reveal solutionSolution

    Energy gained by a charge accelerated through a potential difference depends only on charge × voltage, not on mass — so it is 1 eV regardless of which particle is being accelerated.

    Work-energy theorem for a charge q accelerated from rest through potential difference V:

    $KE = qV$

    For an elementary charge $q = e = 1.6\times10^{-19}\ \text{C}$ and $V = 1\ \text{V}$:

    $KE = (1.6\times10^{-19})(1) = 1.6\times10^{-19}\ \text{J} = 1\ \text{eV}$ (by definition of the electron-volt)

    This result does not involve mass at all — the 1837 factor (proton-to-electron mass ratio) is a distractor; it would only matter if the question asked for the final speed or momentum, not the kinetic energy.

    ✓Final answer

    The correct option is (b) 1 eV.

12. CBSE 2025Set ANNUAL1 markMCQ
    Q.Electric potential 'V' at a distance 'r' from a point charge is directly proportional to ______.

    (a) r

    (b) r²

    (c) 1/r

    (d) 1/r²

    ›Reveal solutionSolution

    The electric potential of a point charge falls off as the inverse of the distance from it.

    The electric potential at a distance $r$ from a point charge $q$ is

    $V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}$

    Since $q$ and $\dfrac{1}{4\pi\varepsilon_0}$ are constants for a given charge, $V \propto \dfrac{1}{r}$. (This is different from the electric field, which varies as $1/r^2$.)

    ✓Final answer

    (c) $1/r$.