Q.Three separate arrangements of vertical equipotential lines are set up in space, the potentials labelled 10 V, 20 V, 30 V, 40 V and 50 V from left to right. In arrangement

Showing the 12 most recent of 20 on this concept.

1. GUJCET 2026Set x1 markMCQ
   Q.The total charge on a uniformly charged spherical shell having radius R is Q. Then electric potential at a distance $r = R/2$ from the centre of the shell ______. (A) $\frac{Q}{4\pi\varepsilon_0 R}$ (B) $\frac{Q}{\pi\varepsilon_0 R}$ (C) $\frac{Q}{2\pi\varepsilon_0 R}$ (D) $\frac{Q}{8\pi\varepsilon_0 R}$

   ›Reveal solutionSolution

   Inside a shell, $V$ is constant $= Q/4\pi\varepsilon_0 R$.

   For a uniformly charged spherical shell, the field inside is zero, so the potential everywhere inside equals its surface value:

   $V(r\le R) = \frac{Q}{4\pi\varepsilon_0 R}.$

   At $r=R/2$ (inside), $V = \dfrac{Q}{4\pi\varepsilon_0 R}$.

   ✓Final answer

   Option (A) $\dfrac{Q}{4\pi\varepsilon_0 R}$

   ANSWER: (A)

2. GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQ
   Q.At ___ distance electric potential due to a electric charge 4 x 10^-7 C will be 4 x 10^4 V.

   (a) 9 m

   (b) 9 micro-m

   (c) 9 cm

   (d) 9 mm

   ›Reveal solutionSolution

   The potential due to a point charge falls off as V = k q / r, so the distance at which a given potential is reached can be found by rearranging for r.

   r = k q / V

   Given q = 4 x 10^-7 C, V = 4 x 10^4 V, k = 9 x 10^9 N m^2/C^2.

   r = (9 x 10^9 x 4 x 10^-7) / (4 x 10^4)

   = (36 x 10^2) / (4 x 10^4)

   = 9 x 10^(2-4)

   = 9 x 10^-2 m = 9 cm.

   ✓Final answer

   (c) 9 cm.

3. GUJCET 2025Set 031 markMCQ
   Q.As shown in figure charges $+q$, $+q$, $-q$ and $-q$ are placed on the vertices of square, each side length is $2l$. The electric potential at mid-point 'A' of charges $+q$ and $+q$ is ______. [FIGURE: a square of side $2l$ with $+q$ at top-left and bottom-left, $-q$ at top-right and bottom-right; A is the midpoint of the left side, between the two $+q$ charges.] (A) Zero (B) $\frac{2kq}{l}\left[1+\frac{1}{\sqrt{5}}\right]$ (C) $\frac{kq}{l}\left[1-\frac{1}{\sqrt{5}}\right]$ (D) $\frac{2kq}{l}\left[1-\frac{1}{\sqrt{5}}\right]$

   ›Reveal solutionSolution

   A is a distance $l$ from each $+q$ and $l\sqrt5$ from each $-q$; $V_A = \frac{2kq}{l}\left(1-\frac{1}{\sqrt5}\right)$.

   A is the midpoint of the left side, so its distance to each left charge $(+q)$ is $l$. Its distance to each right charge $(-q)$: horizontal $2l$, vertical $l$, so $\sqrt{(2l)^2+l^2}=l\sqrt5$.

   $V_A = k\left[\dfrac{q}{l}+\dfrac{q}{l}-\dfrac{q}{l\sqrt5}-\dfrac{q}{l\sqrt5}\right] = \dfrac{2kq}{l}\left(1-\dfrac{1}{\sqrt5}\right).$

   ✓Final answer

   $V_A = \dfrac{2kq}{l}\left[1-\dfrac{1}{\sqrt5}\right].$

   ANSWER: (D)

   

4. GUJCET 2025Set 031 markMCQ
   Q.Charge $1.6\times10^{-7}$ C are distributed uniformly over the surface of spherical conductor of radius R. The ratio of electric potential inside the spherical conductor to the electric field on the surface is ______. (A) $1.6\times10^{-7}R^{2}$ (B) R (C) $1.6\times10^{-7}R$ (D) $\frac{1}{R}$

   ›Reveal solutionSolution

   [!TLDR]

   Inside potential $= kQ/R$ and surface field $= kQ/R^2$, so their ratio is $R$.

   Concept

   For a charged conducting sphere, the potential is constant everywhere inside and equal to its surface value $V = \dfrac{kQ}{R}$; the field just outside the surface is $E = \dfrac{kQ}{R^{2}}$.

   Solution

   $\frac{V_{\text{inside}}}{E_{\text{surface}}} = \frac{\dfrac{kQ}{R}}{\dfrac{kQ}{R^{2}}} = \frac{kQ}{R}\times\frac{R^{2}}{kQ} = R.$

   The charge cancels, so the ratio is simply $R$.

   [!ANSWER]

   (B) R

5. GUJCET 2025Set 031 markMCQ
   Q.The potential difference between two plates of parallel plate capacitor is 2V. As shown in figure electrons are placed at point P and Q. So [FIGURE: a parallel plate capacitor with a positively charged top plate and negatively charged bottom plate, potential difference 2V; electron at point P near the top plate and electron at point Q near the middle/lower region.] (A) Electric forces acting on both the electrons are same. (B) Electric force acting on the electron at point P is greater than the electron at point Q. (C) Electric force acting on the electron at point P is less than the electron at point Q. (D) Electric forces acting on both the electrons are zero.

   ›Reveal solutionSolution

   [!TLDR] Both P and Q lie between the plates of a parallel-plate capacitor, where the field is uniform, so the electric force (F = eE) on each electron is identical regardless of position.

   The field between the plates of a parallel plate capacitor is uniform: E = V/d, the same magnitude and direction at every point in the gap. It does not depend on how close a point is to either plate. The figure shows both P and Q located between the plates. The force on an electron is F = eE. Since E is the same at P and at Q, the electric force on the electron at P equals the force on the electron at Q. The forces are non-zero (there is a field) and equal.

   [!ANSWER] The electric forces on both electrons are equal because the field between the plates is uniform.

   ANSWER: (A)

   

6. GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQ
   Q.Calculate the potential at a point P due to a charge of 4 x 10^-7 C located 9 cm away.

   (a) 4 x 10^4 V

   (b) 4 x 10^-4 V

   (c) 4 x 10^5 V

   (d) 4 x 10^3 V

   ›Reveal solutionSolution

   The electrostatic potential due to a point charge at distance r is V = kq/r.

   V = (9 × 10⁹)(4 × 10⁻⁷)/(0.09) = 3600/0.09 = 4 × 10⁴ V.

   ✓Final answer

   (a) 4 × 10⁴ V.

7. GUJCET 2024Set 131 markMCQ
   Q.If an electron is accelerated by a potential difference of 2.5 V it would gain energy of ________. (Take charge of electron $1 \times 10^{-19}$ C) (A) 2.5 erg (B) 2.5 MeV (C) 2.5 eV (D) 2.5 J

   ›Reveal solutionSolution

   Accelerating an electron through $2.5$ V gives it energy $qV = e\times2.5\,\text{V} = 2.5\,\text{eV}$ by definition of the electron-volt.

   Concept. The energy gained by charge $q$ across potential difference $V$ is $W=qV$. For one electronic charge across $1$ V the energy is exactly $1$ eV.

   Steps. Here $V=2.5$ V, so the electron gains $2.5$ eV.

   ✓Final answer

   (C) 2.5 eV

   ANSWER: (C)

8. GUJCET 2024Set 131 markMCQ
   Q.A radius of spherical charged shell is 10 cm and electric potential on its surface is 100 V, then the potential at 2 cm from the centre of the shell will be ________. (A) 0 V (B) 1 V (C) 200 V (D) 100 V

   ›Reveal solutionSolution

   Inside a charged spherical shell the potential is uniform and equal to its surface value, so at $2$ cm (inside the $10$ cm shell) it is $100$ V.

   Concept. For a charged conducting/spherical shell, the field inside is zero and the potential everywhere inside equals the surface potential.

   Steps. Since $2\,\text{cm} < 10\,\text{cm}$ (radius), the point is inside, so $V = V_{\text{surface}} = 100$ V.

   ✓Final answer

   (D) 100 V

   ANSWER: (D)

9. GUJCET 2023Set 091 markMCQ
   Q.A charge Q is placed at the centre of circle of radius 10 cm. Find the work done in moving a charge q between any two points lying on the arc of this circle. (A) KQq J (B) 0.1 KQq J (C) 0.5 KQq J (D) 0 J

   ›Reveal solutionSolution

   [!TLDR]

   Points on the arc are equidistant from Q, hence equipotential, so the work done is zero.

   Concept

   The work done in moving a charge $q$ between two points equals $q$ times the potential difference: $W = q(V_A - V_B)$. On an equipotential surface this difference is zero.

   Solution

   The potential due to Q at distance $r$ is $V = \dfrac{KQ}{r}$, which depends only on $r$. Every point on the arc of the circle is at the same distance ($r = 10$ cm) from the central charge Q, so all such points have the same potential. Therefore the potential difference between any two arc points is zero, and

   $W = q \Delta V = 0.$

   [!ANSWER]

   (D) 0 J

10. GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQ
    Q.Figure shows the field lines of a positive and negative charge respectively. Give the sign of potential difference V_Q - V_P, V_B - V_A.

    (a) -ve, +ve

    (b) +ve, -ve

    (c) +ve, +ve

    (d) -ve, -ve

    ›Reveal solutionSolution

    Near a positive charge the closer point has higher potential (V_Q - V_P > 0); near a negative charge the farther point has the higher (less negative) potential (V_B - V_A > 0). Both are positive.

    Potential of a point charge: V = kq/r.

    Positive charge (left): Q is nearer, P is farther. V decreases with r, so V_Q > V_P:

    V_Q - V_P = positive (+ve).

    Negative charge (right): V = -k|q|/r. A is nearer (more negative), B is farther (less negative, i.e. higher). So V_B > V_A:

    V_B - V_A = positive (+ve).

    Both potential differences are positive.

    ✓Final answer

    (c) +ve, +ve.

11. GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQ
    Q.The electric potential energy of 2 microC charge is 3 x 10^-5 J at a point in a uniform electric field. The electric potential at that point is ___ V.

    (a) 5

    (b) 15

    (c) 6

    (d) Zero

    ›Reveal solutionSolution

    Electric potential = potential energy per unit charge, V = U/q = 3x10^-5/2x10^-6 = 15 V.

    Electric potential is the potential energy per unit charge:

    V = U/q = (3 x 10^-5 J)/(2 x 10^-6 C) = 15 V.

    ✓Final answer

    (b) 15 V.

12. GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQ
    Q.Equipotential surfaces at a very large distance from the collection of charges whose total sum is not zero are approximately ___.

    (a) paraboloids

    (b) planes

    (c) spheres

    (d) ellipsoid

    ›Reveal solutionSolution

    At large distances a collection of charges with nonzero total behaves like a single point charge, so its equipotential surfaces become concentric spheres.

    Very far from a group of charges whose net charge is not zero, the potential is dominated by the monopole term V approximately k Q_total/r - exactly like that of a point charge.

    Surfaces of constant potential (r = constant) are therefore concentric spheres.

    ✓Final answer

    (c) spheres.