NCERT Exemplar Solutions for CBSE Class 12 Physics, Chapter 2 (Part I)
Get NCERT Exemplar Solutions for Class 12 Physics, Chapter 2: Electrostatic Potential and Capacitance. In this article, you will get solutions from question number 2.1 to 2.6. These are basically multiple choice questions with a single correct option. All questions are important for CBSE 12th board exam 2017.
NCERT Exemplar Solutions for Class 12 Physics, Chapter 2: Electrostatic Potential and Capacitance are available here. Here you will get solutions for multiple choice correct answer i.e., from question number 2.1 to 2.6. Solutions of other questions are available in other parts. Questions from NCERT Exemplar are frequently asked in CBSE board exams and other competitive exams like JEE Main, JEE Advanced, NEET etc.
Questions and solutions are given below:
Question 2.1: A capacitor of 4 μ F is connected as shown in the circuit (Fig. 2.1). The internal resistance of the battery is 0.5 Ω. The amount of charge on the capacitor plates will be
(b) 4 μ C
(c) 16 μ C
(d) 8 μ C
The capacitor offers infinite resistance in a DC circuit. Here, no current will flow through the capacitor and 10Ω resistance after the capacitor is full charged.
The potential difference across 10Ω resistance will be zero. It will act like a plain wire.
Current flowing through 2Ω resistance is given by
I = V/(R + r) = 2.5/(2 + 0.5) = 1 A
Potential difference across 2Ω resistance V = I R = 1 × 2 = 2V
Here, capacitor is connected in parallel with 2 Ω resistance, so it will also have 2V potential difference across it.
The charge on capacitor, q = CV = (2 mF) × 2V = 8 m C.
Question 2.2: A positively charged particle is released from rest in an uniform electric field. The electric potential energy of the charge
(a) remains a constant because the electric field is uniform.
(b) increases because the charge moves along the electric field.
(c) decreases because the charge moves along the electric field.
(d) decreases because the charge moves opposite to the electric field.
The positive charge will experience an electrostatic force whose direction will be along the direction of electric field.
In other words, positive charge will move from high electrostatic potential to low electrostatic potential.
Work will be done by electric filed on the charge and the electric potential energy of the charge will decrease.
Question 2.3: Figure 2.2 shows some equipotential lines distributed in space. A charged object is moved from point A to point B.
a) The work done in Fig. (i) is the greatest.
(b) The work done in Fig. (ii) is least.
(c) The work done is the same in Fig. (i), Fig. (ii) and Fig. (iii).
(d) The work done in Fig. (iii) is greater than Fig. (ii)but equal to that in Fig. (i).
Work done in moving a charge q from point A to point B is given by WAB = q (VB ‒VA).
Here, values of VA and VB are same in all the cases. So, work done is the same in Fig. (i), Fig. (ii) and Fig. (iii).
Question 2.4: The electrostatic potential on the surface of a charged conducting sphere is 100V. Two statments are made in this regard:
S1: At any point inside the sphere, electric intensity is zero.
S2: At any point inside the sphere, the electrostatic potential is
Which of the following is a correct statement?
(a) S1 is true but S2 is false
(b) Both S1 & S2 are false
(c) S1 is true, S2 is also true and S1 is the cause of S2
(d) S1 is true, S2 is also true but the statements are independent
Electric field intensity E and electric potential V are related as E = ‒ (dV/dR).
Here, E = 0 ⇒ (dV/dr) = 0.
Now, (dV/dr) = 0 when V is constant (or V = 0).
Question 2.5: Equipotentials at a great distance from a collection of charges whose total sum is not zero are approximately
Total sum of collection of charges is not zero it means there is some net positive or negative charge.
At a very large distance this collection of charge will act as a point charge.
So, equipotentials at a very large distance from this point charge will be spheres.
Question 2.6: A parallel plate capacitor is made of two dielectric blocks in series. One of the blocks has thickness d1 and dielectric constant k1 and the other has thickness d2 and dielectric constant k2 as shown in Fig. 2.3.
This arrangement can be thought as a dielectric slab of thickness d (= d1+d2) and effective dielectric constant k. The k is
Capacitance C, of parallel plate capacitor filled with dielectric block has thickness d and dielectric constant K is given by C = (K εo A)/d, where A is area of the plate.
Now, the capacitance of parallel plate capacitor filled with dielectric block has thickness d1 and dielectric constant K1 is given by C1 = (K1εo A)/d1,
The capacitance of parallel plate capacitor filled with dielectric block has thickness d2 and dielectric constant K2 is given by C2 = (K2εo A)/d2,
The situation shown in figure is similar to the combination of two capacitors connected in series, the equivalent capacitance is given by C = (K εo A)/d,
(1/C) = (1/C1) + (1/C2)
⇒ C = C1C2/(C1+C2)
Putting values of C, C1 and C2 and solving we will get
K = [K1K2(d1 + d2)]/[K1d2 + K2d1]