NIOS Class 12 Physics Chapter 16 Electric Potential and Capacitors

NIOS Class 12 Physics Chapter 16 Electric Potential and Capacitors Solutions English Medium As Per New Syllabus to each chapter is provided in the list so that you can easily browse throughout different chapters NIOS Class 12 Physics Chapter 16 Electric Potential and Capacitors Notes in English and select need one. NIOS Class 12 Physics Solutions English Medium Download PDF. NIOS Study Material of Class 12 Physics Notes Paper Code: 312.

NIOS Class 12 Physics Chapter 16 Electric Potential and Capacitors

Also, you can read the NIOS book online in these sections Solutions by Expert Teachers as per National Institute of Open Schooling (NIOS) Book guidelines. These solutions are part of NIOS All Subject Solutions. Here we have given NIOS Class 12 Physics Notes, NIOS Senior Secondary Course Physics Solutions in English for All Chapter, You can practice these here.

Electric Potential and Capacitors

Chapter: 16

Module-V: Electricity and Magnetism

INTEXT QUESTIONS 16.1

1. A metallic sphere of radius R has a charge +q uniformly distributed on its surface. What is the potential at a point r ( > R) from the centre of the sphere?

Ans: For a metallic sphere with charge +q uniformly distributed on its surface, we treat it as a point charge located at the center when finding potential at external points.

Using the formula for potential due to a point charge:

V = (1/4πε₀) × (q/r)

V = q/(4πε₀r)

2. Calculate the work done when a point charge is moved in a circle of radius r around a point charge q.

Ans: When a charge is moved along an equipotential surface (such as a circle around a point charge), no work is done because:

(i) All points on the circle are equidistant from the central charge.

(ii) Hence, all points have the same potential.

(iii) Work done = q₀(V_final – V_initial) = q₀(V – V) = 0

: Work done = 0

3. The electric potential V is constant in a region. What can you say about the electric field E in this region?

Ans: 

In simpler terms, the electric field measures how rapidly the potential changes with distance. If the potential does not change, the electric field must be zero. Therefore, in a region where the electric potential remains constant, the electric field is absent or zero everywhere.

4. If electric field is zero at a point, will the electric potential be necessarily zero at that point?

Ans: No, electric potential need not be zero when electric field is zero.

The reasons are the following:

(i) Electric field E = -dV/dr = 0 means dV/dr = 0

(ii) This indicates that potential is not changing at that point (potential is at maximum, minimum, or constant)

(iii) But the potential value itself can be any finite value, not necessarily zero

(iv) Example: At the center between two equal positive charges, E = 0 but V ≠ 0

: No, electric potential need not be zero when electric field is zero at that point.

5. Can two equipotential surfaces intersect?

Ans: No, two equipotential surfaces cannot intersect.

The reasons are the following:

(i) If two equipotential surfaces intersect, there would be two different potential values at the point of intersection.

(ii) This is impossible because potential at any point has a unique value.

(iii) Also, electric field is always perpendicular to equipotential surfaces.

(iv) If two surfaces intersect, electric field would have two different directions at the same point, which is impossible: No, two equipotential surfaces cannot intersect.

INTEXT QUESTIONS 16.2

1. Write the dimensions of capacitance.

Ans: From C = Q/V:

Dimensions of Q (charge) = [A T]

Dimensions of V (potential) = [M L² T⁻³ A⁻¹]

Dimensions of C = [A T] / [M L² T⁻³ A⁻¹] 

= [M⁻¹ L⁻² T⁴ A²]

: [M⁻¹ L⁻² T⁴ A²]

2. What is the potential difference between two points separated by a distance d in a uniform electric field E?

Ans: Following changes occur in the related quantities:

(i) Capacitance (C): The capacitance increases by a factor of K.
C = K × C₀

(ii) Electric Field (E): The electric field decreases by a factor of K.
E = E₀ / K

(iii) Potential Difference (V): The potential difference also decreases by a factor of K (if charge remains constant).
V = V₀ / K

So, when a dielectric is introduced:

Capacitance increases.

Electric field and potential difference decrease.

All changes are directly related to the dielectric constant K.