Home >> NCERT Solutions >> Class 12 – Physics >> Chapter 2 – Electrostatic Potential and Capacitance
NCERT Solutions Class 12 Physics Chapter 2 – Electrostatic Potential and Capacitance
2.1
2.2
2.3
Ans – (a) Two charges are represented by,
A plane of any kind with identical electric potential across all points can be designated as an equipotential surface. Parallel to line AB constitutes a single plane. Due to the identical magnitude of these electrical charges, the surface is positioned near the middle of the horizontal line AB.
(b) We may state that the electric field will have an impact perpendicular to the plane present within the axis of AB at each position along this surface, as seen in the above image. Since we are all aware that both force & field operate perpendicular to one another. This indicates that the space available will just extend in that region since we are already close to the equipotential field.
2.4
Ans – (a) As we all know from the above equation,
The spherical conductor’s radius is represented as, r = 12cm = 0.12m
The electrical conductor has an equal distribution of the load. Since there is no electrical charge in the material that conducts overall, there is no electric field present in a cylindrical conductor.
(b)
(c) Let electric field at a point 18 cm from the centre of the sphere be E1.
Distance of given point from the centre, d = 18 cm = 0.18m
2.5
Ans – Given that,
Capacitance, C = 8 pF
Distance between parallel plates = d
Air was filled in it, ⇒ dielectric constant of air, k = 1
Now,
Distance between plates is decreased to half, ⇒ d‘ = d/2
Dielectric constant of substance filled in between plates is 6, ⇒ k‘ = 6
2.6
Ans – (a)
(b) Given that,
V = 120 V
The potential difference V‘ across each capacitor will equal one-third of the supply voltage.
V‘ = 120/3 = 40 V
2.7
Ans – (a) Given C1 = 2pF, C2 = 3pF, C3 = 4pF
Formula for equivalent capacitance (C′) of capacitors’ parallel combination is given by
C′ = C1 + C2 + C3
⇒ C′ = 2 + 3 + 4 = 9pF
(b) Given supply voltage V=100 V
Charge on a capacitor with capacitance C and potential difference V is given by,
q = CV ……(i)
For C = 2pF,
q = 100 × 2 = 200pF
For C = 3pF,
q = 100 × 3 = 300pF
For C = 4pF,
q = 100 × 4 = 400pF
2.8
Ans – Given that,
Area of each plate, A = 6 × 10−3 m2
Distance separating the plates, d = 3mm = 3 × 10−3m
Supply voltage, V = 100 V
2.9
Ans – (a) Given that,
Mica sheet’s Dielectric constant, k = 6
Initial capacitance, C = 17.71 × 10−12 F
New capacitance, C′ = kC
⇒ C′ = 6 × 17.71×10−12 = 106 pF
Supply voltage, V = 100V
New charge, q′ = C′V′
⇒ q′ = 106 × 100 pC = 1.06 × 10−8 C
Potential across plates will remain 100 V.
(b) Given that,
2.10
2.11
Class 12 Physics Chapter-2 Electrostatic Potential And Capacitance Overview
Electrostatic Potential and Capacitance. It is easy to grasp how charges create potential and how energy is stored in electric fields with NCERT Solutions. First, the chapter talks about potential difference, electric field, and the work done when charges move. These are ideas that come up again in electromagnetism.
Also, real things like pacemakers and camera flashes require capacitors. But a lot of students don’t get the math. That’s why the Electrostatic Potential and Capacitance NCERT Solutions have clear instructions, illustrations, and examples that are easy to understand. This makes learning faster and easier.
The NCERT update in 2025 got rid of repeating theory and introduced issues about dielectrics, energy density, and V–Q graphs that happen in real life. Because of this, students now pay greater attention to thinking and application. You will always be on the same page as the syllabus because our content covers all of these changes.
These Electrostatic Potential and Capacitance NCERT Solutions are ready for tests, focus on the CBSE, and are easy for students to understand. The solved examples and review advice can help you get faster and more confident, whether you’re studying for the boards or the JEE.
FAQs –Electrostatic Potential And Capacitance Class 12 Chapter-2 NCERT
Potential is energy per unit charge; think of it like “height” in a gravity analogy. Our solutions repeatedly label units to cement the difference.
Remember, moving against the field raises potential (positive work), while moving with it lowers potential (negative). Each example shows sign decisions step-by-step.
If plates share the same voltage they’re in parallel; otherwise, they’re in series. Flow-chart summaries in the solutions reinforce this decision quickly.
The new NCERT dropped repetitive derivations and instead added dielectrics in spherical capacitors, energy density proofs, and more graph-based questions.
Therefore, each problem ends with a boxed answer and a units check. Cross-reference those while practicing to spot slips instantly.