Electric Potential Numerical on Potential Gradient Class-12 Nootan ISC Physics Solutions Ch-3. Step by step solutions of Kumar and Mittal Physics of Nageen Prakashan as council latest prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ISC Board Class-12 Physics.
Electric Potential Numerical on Potential Gradient Class-12 Nootan ISC Physics Solutions Ch-3
| Board | ISC |
| Class | 12 |
| Subject | Physics |
| Book | Nootan |
| Chapter-3 | Electric Potential |
| Topics | Numerical on Potential Gradient |
| Academic Session | 2025-2026 |
Numerical on Potential Gradient
Que-18: 10 J of work is required to move a charge of – 200 C from point A to B. (i) Which of the two points is at higher potential? (ii) What is the PD between them? (iii) If the points are 4 cm apart, what is the component of the electric field parallel to the line joining A and B?
Ans-18: W = 10 J
q = -200 C
(i) W = (VB-VA)q
10 = (VB-VA)(-200)
(VB-VA) = 10/-200 = -1/20 V
VA > VB.
(ii) PD between them
VA – VB = 1/20 = 0.05 V.
(iii) E.F = V/r
=0.05/(4 x 10‾²)
= 5/4 = 1.25 V/m.
Que-19: 3.24 x 10^4 J of work is done in taking 3.0 × 10^-8 C of charge in a uniform electric field from one point to the other. If the distance between the points be 12 cm, calculate the intensity of the electric field.
Ans-19: Given : W = 3.24 x 10^4 J
q = 3.0 x 10^-8 C
r = 12cm = 12 x 10^-2 m
V = W/q
= (3.24 x 10^4)/(3 x 10^-8)
= 1.08 x 10^4 V
E = V/r
= (1.08 x 10^4)/(12 x 10^-2)
= 9 x 10^4 V/m.
Que-20: Two plane metallic plates are at a distance of 2 cm. These are connected to a battery of 1000 V. A proton is placed between these plates. Find out (i) intensity of electric field between the plates, (ii) force on the proton. Will the proton experience different forces at different places between the two plates?
Ans-20:
(i) E = V/r
= 1000/(2 x 10^-2)
= 50000 V/m.
(ii) F = qE
= 1.6 10^-19 x 50000
= 8.0 x 10^-15 N.
Que-21: The distance between two horizontal parallel plates is 2.0 cm and the potential difference between them is 120 V. Calculate: (i) electric field between the plates, (ii) force on an electron passing in between the plates, (iii) increase in electrical energy of the electron in moving from one plate to the other.
Ans-21: Given : V = 120 V
r = 2cm = 2 x 10^-2 m
(i) E = V/r
= 120/(2 x 10^-2)
= 6.0 x 10³ V/m.
(ii) F = qE
= 1.6 x 10^-19 x 6.0 x 10³
= 9.6 x 10^-16 N.
(iii) Increase in energy = work done
I = eV
= 1.6 x 10^-12 x 120
= 1.92 x 10^-17 J.
— : End Gauss Theorem Numerical on Charged Spherical Shell Class-12 Nootan ISC Physics Ch-2. :–
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