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PhysicsMediumClass 12

Equipotential Surfaces

Electrostatics

18

JEE Qs

8%

Hard

60

min

Master the perpendicularity of electric field lines to equipotential surfaces and the zero work done property, as these are critical for solving conceptual and problem-based questions.

🧮 Key Formulas

V = constant (on an equipotential surface)
W_AB = q(V_B - V_A)
If V_B = V_A, then W_AB = 0
E = -dV/dr (magnitude in 1D)
E = -∇V (general vector relation)
∫(E ⋅ dl) = 0 (along an equipotential surface)

✅ Key Points for JEE

  • 1An equipotential surface is a locus of points where the electric potential is constant. This means the potential difference between any two points on the surface is zero.
  • 2No work is done by the electric field when a charge is moved along an equipotential surface. Therefore, the tangent to an equipotential surface has no component of the electric field.
  • 3Electric field lines are always perpendicular to equipotential surfaces at every point. The electric field points in the direction of decreasing potential.
  • 4Equipotential surfaces are closer together in regions where the electric field is stronger (i.e., where potential changes rapidly with distance) and farther apart where the field is weaker.
  • 5No two equipotential surfaces can intersect, as intersection would imply two different values of potential at the same point, which is physically impossible.

⚠️ Common Mistakes

  • Confusing electric field lines with equipotential surfaces, or assuming they are parallel. Remember, they are always perpendicular.
  • Incorrectly assuming that any work done by an external agent moving a charge on an equipotential surface is zero; only the work done by the *electric field* is zero.
  • Failing to visualize equipotential surfaces in 3D (e.g., concentric spheres for a point charge) and only thinking of them as 2D lines.

📝 Practice Questions

See all

Q38.A capacitor, C1 = 6μ F is charged to a potential difference of V0 = 5 V using a 5 V battery. The battery is removed and another capacitor, C2 = 12μ F is inserted in place of the battery. When the switch 'S' is closed, the charge flows between the capacitors for some time until equilibrium condition is reached. What are the 2025 (29 Jan Shift 2) JEE Main Previous Year Paper charges (q1 and q2) on the capacitors C1 and C2 when equilibrium condition is reached. (1) q1 = 10μC, q2 = 20μC (2) q1 = 30μC, q2 = 15μC (3) q1 = 20μC, q2 = 10μC (4) q1 = 15μC, q2 = 30μC

2025·NumericalMedium

Q44.A point charge causes an electric flux of −2 × 104Nm2C−1 to pass through a spherical Gaussian surface of 8.0 cm radius, centred on the charge. The value of the point charge is : (Given ϵ0 = 8.85 × 10−12C2 N−1 m−2 ) (1) 15.7 × 10−8C (2) 17.7 × 10−8C (3) −15.7 × 10−8C (4) −17.7 × 10−8C

2025·MCQEasy

Q45.An electric dipole is placed at a distance of 2 cm from an infinite plane sheet having positive charge density σo . Choose the correct option from the following. (1) Potential energy and torque both are maximum. (2) Torque on dipole is zero and net force is directed away from the sheet. (3) Torque on dipole is zero and net force acts (4) Potential energy of dipole is minimum and torque towards the sheet. is zero.

2025·MCQMedium

Q26.An electron is made to enter symmetrically between two parallel and equally but oppositely charged metal plates, each of 10 cm length. The electron emerges out of the electric field region with a horizontal component of velocity 106 m/s. If the magnitude of the electric field between the plates is 9.1 V/cm , then the vertical component of velocity of electron is (mass of electron = 9.1 × 10−31 kg and charge of electron = 1.6 × 10−19C ) (1) 0 (2) 1 × 106 m/s (3) 16 × 106 m/s (4) 16 × 104 m/s

2025·MCQMedium

Q38.A line charge of length ' a ' is kept at the center of an edge BC of a cube ABCDEFGH having edge length ' 2 a ' as shown in the figure. If the density of line charge is λ C per unit length, then the total electric flux through all the faces of the cube will be . (Take, ϵ0 as the free space permittivity) (1) λa (2) λa 2ϵ0 4ϵ0 (3) λa (4) λa 16ϵ0 8ϵ0

2025·MCQHard

Q44.A parallel-plate capacitor of capacitance 40μ F is connected to a 100 V power supply. Now the intermediate space between the plates is filled with a dielectric material of dielectric constant K = 2. Due to the introduction of dielectric material, the extra charge and the change in the electrostatic energy in the capacitor, respectively, are (1) 4 mC and 0.2 J (2) 8 mC and 2.0 J (3) 2 mC and 0.4 J (4) 2 mC and 0.2 J 2025 (22 Jan Shift 1) JEE Main Previous Year Paper

2025·MCQMedium

NCERT Chapters

  • Class 12 Physics Ch 2: Electrostatic Potential and Capacitance