Question 1
Charges +5×10⁻⁸ C and −3×10⁻⁸ C are 16 cm apart. Find the points on their line where potential is zero.
Steps only
5/x=3/(16−x) ⇒ x=10 cm; 5/x=3/(x−16) ⇒ x=40 cm
Answer
At x=10 cm and x=40 cm from + charge
CLASS 12 · PHYSICS · CHAPTER 2
Electrostatic potential and capacitance
Easy, worked answers to all 11 current exercises on electric potential, fields, capacitors and stored energy.
11 worked answersComplete chapter answers
11 worked answers · short steps first, explanations when you need them
Chapter 2 Exercises
Textbook pages 79–80
Question 2
Each vertex of a regular hexagon of side 10 cm has charge +5 μC. Find potential at its centre.
Steps only
V=6kq/r=6(9×10⁹)(5×10⁻⁶)/0.10=2.7×10⁶ V
Answer
2.7 MV
Question 3
Equal and opposite charges are 6 cm apart. Identify an equipotential surface and the electric-field direction on it.
Steps only
On the perpendicular bisector, distances to +q and −q are equal, so potentials cancel; E points from + to −.
Answer
Midpoint perpendicular plane; E from + toward −
Question 4
A conducting sphere has radius 12 cm and charge 1.6×10⁻⁷ C. Find E (a) inside, (b) just outside, and (c) 18 cm from its centre.
Steps only
Inside conductor E=0; outside E=kq/r²: 1.0×10⁵ N/C at 0.12 m and 4.4×10⁴ N/C at 0.18 m.
Answer
0; 1.0×10⁵ N/C; 4.4×10⁴ N/C
Question 5
An air capacitor is 8 pF. Find its capacitance if plate spacing is halved and a dielectric of constant 6 fills the gap.
Steps only
Cnew=Cold×2×6=8×12=96 pF
Answer
96 pF
Question 6
Three 9 pF capacitors are in series across 120 V. Find equivalent capacitance and voltage across each.
Steps only
Ceq=9/3=3 pF; Veach=120/3=40 V
Answer
3 pF; 40 V each
Question 7
Capacitors 2 pF, 3 pF and 4 pF are in parallel across 100 V. Find total capacitance and charge on each.
Steps only
Ceq=9 pF; Qᵢ=CᵢV gives 200, 300 and 400 pC.
Answer
9 pF; 200 pC, 300 pC, 400 pC
Question 8
An air capacitor has plate area 6×10⁻³ m² and gap 3 mm. Find capacitance and charge at 100 V.
Steps only
C=ε₀A/d=17.7 pF; Q=CV=1.77 nC
Answer
17.7 pF; 1.77 nC
Question 9
For the capacitor in Exercise 2.8, a 3 mm mica sheet with K=6 fills the gap. What happens (a) with the battery connected and (b) after it is disconnected?
Steps only
C increases sixfold. Connected: V fixed, so Q increases sixfold. Disconnected: Q fixed, so V falls sixfold.
Answer
Connected: 10.6 nC at 100 V; disconnected: 1.77 nC at 16.7 V.
Question 10
Find the energy stored in a 12 pF capacitor connected to a 50 V battery.
Steps only
U=½CV²=½(12×10⁻¹²)(50²)=15×10⁻⁹ J
Answer
15 nJ
Question 11
A 600 pF capacitor charged to 200 V is disconnected and connected to an identical uncharged capacitor. Find the energy lost.
Steps only
Ui=12 μJ; final voltage=100 V; Uf=6 μJ; loss=Ui−Uf=6 μJ
Answer
6 μJ lost