Electric Charge and Fields Numerical on Electric Field and Intensity Class-12 Nootan ISC Physics Nootan Solutions Ch-1. 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 Charge and Fields Numerical on Electric Field and Intensity Class-12 Nootan ISC Physics Nootan Solutions Ch-1
| Board | ISC |
| Class | 12 |
| Subject | Physics |
| Book | Nootan |
| Chapter-1 | Electric Charge and Fields Numerical |
| Topics | Numericals on Electric Field and Intensity |
| Academic Session | 2025-2026 |
Solved Numericals on Electric Field Lines and Intensity
Electric Charge and Fields Numerical on Electric Field and Intensity Class-12 Nootan ISC Physics Nootan Solutions Ch-1
Que-19: An alpha particle is placed in an electric field of 15 x 10^4 N/C. Calculate the force on the particle.
Ans-19: F = qE
= 3.2 x 10^-19 x 15 x 10^4
= 4.8 x 10^-14 N
Que-20: A body having an excess of 10^6 electrons is placed in an electric field of 1000 N/C towards East. Find out the magnitude and direction of the force acting on the body.
Ans-20: Body has charge
Q = ne
= -1.6 x 10^-13 x 1000
= -1.6 x 10^10 N
negative sign shows that force is opposite to field i.e. due west
Que-21: What will be the intensity of an electric field in which an electron experiences an electric force equal to its weight? What in case of proton?
Ans-21: given F = mg
= 9 x 10^-31 x 9.8 N
now F = qE
Therefore, E = F/q
= 9 x 9.8 x 10^-31 / 1.6 x 10^-19
= 5.5 x 10^-11 N C
electron experiences force opposite to field
Therefore field is directive downward
again in case of proton
E = F/q = mg/q
= 1.7 x 10^-27 x 9.8 / 1.6 x 10^-19
= 1.04 x 10^-7 N/C vertically upward.
Que-22: Calculate the intensity of the electric field due to a helium nucleus at a distance of 1 Å from the nucleus.
Ans-22: q = 3.2 x 10^-19 C
r = 1Å = 1 x 10^-10 m
According to formula
E = 1/4πε0 q/r
= 9 x 10^9 x 3.2 x 10^-19 / (10^-10)^2
= 28.8 x 10^10 N/C or 2.88 x 10^11 N/C
Que-23: Two point-charges q1 = + 0.2 C and q2 = + 0.4 C are 0.1 m apart. Find the electric field at (i) mid-point between the charges, (ii) a point on the line joining q1 and q2 such that it is 0.05 m away from q2 and 0.15 m away from q1.
Ans-23: Case 1
Field at mid point
= 1/4πε0 [0.4 / (0.05)^2 – 0.2 / (0.05)^2]
= 9 x 10^9 x 0.2 / (0.05)^2
= (1.8 / 25 x 10^-4) x 10^9
= 7.2 x 10^11 N/C
In second case
at P the field due to both charges will be in same direction i.e.
E = 1/4πε0 [0.2 / (0.15)^2 + 0.4 / (0.01)^2]
= 9 x 10^9 x 0.2 / (0.05)^2 [1/9 + 2]
= 3.8 x 10^9 / 25 x 10^-4 = 1.52 x 10^12 N/C
charges are in coulomb so 10^-6 is removed
Que-24: Electric charges ± 1000 µC are placed at points A and B, respectively at a distance of 2 m from each other. Calculate the electric fields at (i) mid-point of the line AB and (ii) at a point at equal distances of 4 m from each charge.
Ans-24: At mid point field due to both charges will be along AB
Therefore Net field
E = 2 x 1/4πε0 [1000 x 10^-6/(1)^2]
= 1.8 x 10^-7 N/C
In second case as shown in fig electric field will be along AB
E = (FB + FB) cos
= 1/4πε0 [1000 x 10^-6 x 2 / 4^2 ] x 1/4
= 2.8 x 10^5 N/C
Que-25: A stationary oil drop between two parallel plates has a charge of 3.2 x 10^-19 C and a weight of 1.6 x 10^-14 N. Find the electric field acting on the drop.
Ans-25: F = qE
E = F/q
= 1.6 x 10^-14 / 3.2 x 10^-19
= 1/2 x 10^5 N/C
= 5 x 10^4 N/C
— : End Electric Charge and Fields Numerical on Electric Field and Intensity Class-12 Nootan ISC Physics Nootan Solutions :–
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