Potential Energy — System of charges
Electrostatics
18
JEE Qs
8%
Hard
60
min
Systematically identify all unique pairs of charges and their separation distances, summing their individual potential energies algebraically, and remember to include external field contributions if applicable, paying close attention to signs.
🧮 Key Formulas
✅ Key Points for JEE
- 1Potential energy is a scalar quantity defined for a system of charges, representing the work done by an external agent to assemble the charges from infinite separation to their current configuration without acceleration.
- 2For a system of multiple charges, calculate the potential energy for every unique pair of charges and sum them algebraically. Be careful not to double-count pairs.
- 3The sign of charges is crucial: like charges (q₁q₂ > 0) contribute positive potential energy, while unlike charges (q₁q₂ < 0) contribute negative potential energy.
- 4If an external electric field is present, the total potential energy of the system includes the mutual interaction potential energy AND the potential energy of each individual charge due to this external field (U_total = U_mutual_interaction + Σ qᵢV_ext(rᵢ)).
- 5The potential energy is path-independent because electrostatic force is conservative. The work done by an external agent to move a charge in an electric field is equal to the change in potential energy (ΔU).
⚠️ Common Mistakes
- ✕Double-counting pairs when summing potential energies for a system of multiple charges (e.g., calculating (q1q2)/r and then (q2q1)/r).
- ✕Incorrectly using the signs of charges in the potential energy calculation, leading to errors in the final algebraic sum.
- ✕Forgetting to include the potential energy of individual charges in an external electric field when such a field is present, only considering mutual interactions.
- ✕Confusing work done by an external agent with work done by the electrostatic field (they are negatives of each other).
📝 Practice Questions
See allQ38.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
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
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.
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
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
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
NCERT Chapters
- Class 12 Physics Part 1, Chapter 2: Electrostatic Potential and Capacitance