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NCERT Questions for Class 12 Physics Chapter 1 – Electric Charges and Fields
In Class 12 Physics, Chapter 1 dives into the fundamentals of Electric Charges and Fields, laying the groundwork for understanding the complex interactions between particles in the field of physics. This chapter is crucial for building a strong foundation in the subject and is the starting point for delving into more advanced topics in later chapters.
Important Questions with Solutions of Class 12 Physics Chapter 1 – Electric Charges and Fields
1) Will the energy between the 2 poles be modified when the dielectric constant surrounding the element increases?
Ans – The dielectric constant of a medium is denoted by k such that the force between the charges in a vacuum (Fv) will be divided by the force between two charges in the medium (Fm).
where k = Fv/Fm = force between the vacuum charges/force between the 2 medium charges.
Fm = Fv/k
From this equation, it’s obvious that when k increases, Fm decreases.
2) A charged rod P entices a rod R. Meanwhile P repels other charged rod. Explain what force created between Q & R.
Ans – When the p rod is negatively charged, it attracts the R rod to be positively charged. As we know p pushes away rod Q due to the negative charge of rod Q. The external force between Q & R would pull them together.
3) An electric dipole placed at an angle of 30° due to a uniform electric field of 104 N/C gets a 9 * 10-26 Nm torque. Explain the dipole moment of the dipole.
Ans – As you know,
θ = 30°
τ = 9 * 10-26 Nm
E = 104 N/C
Dipole moment P can be added.
Torque can be found with the expression τ = PEsinθ
Now,
P = τ / Esinθ
Hence, P = 18 * 10-30 Cm
4) a) Mention what the below statement means, “Electric charge of a body is quantized.”
Ans – The statement ‘The electric charge of a body is quantized,’ means that the whole number (1, 2, 3, 4,., n) of electrons can flow from one body to another. It’s understood that charges cannot be passed in parts. So, a body has its total charge in whole numbers of electric charges.
b) Why do people ignore the electric charge quantization while dealing with large-scale charges?
Ans – When dealing with large-scale charges, the number of charges is massive compared to the size of each electric charge. Meanwhile, on a macroscopic scale, the electric charge quantization is useless. Thus, it’s forgotten & considered that electric charge is continuous.
5) An electric dipole is placed at an angle of 30° with the direction of a uniform electric field with magnitude 5 * 104 NC–1 & has a dipole moment of 4 * 10–9 Cm. What will be the magnitude of the torque placed on the dipole?
Ans – As you know,
Electric dipole moment, p = 4 * 10-9 Cm
The uniform electric field makes an angle θ = 30° with p.
Electric field, E = 5 * 104 NC-1
The torque on the dipole is denoted as τ = pEsinθ
τ = 4 * 10-9 * 5 * 104 * sin30°
⇒ τ = 20 * 10-5 * 1/2
Therefore, τ =10-4 Nm
Hence, the torque of magnitude on the dipole will be 10-4 Nm.
6) A particle with a mass of m & a charge of q is set free from others in a uniform electric field of intensity E. Find out the kinetic energy gained by this element while moving a certain distance between plates.
Ans – Electrostatic force on a charge q in an electric field E is denoted as:
F=qE …(1)
We also know Newton’s second law of motion, which states that
F=ma …(2)
From equation (1) & (2),
a = qE / m …(3)
We have the 3rd equation of motion, mentioned as: v2 – u2 = 2as
Because the charged particle starts from rest, u = 0
v2 = 2as …(4)
Now, We have the equation for kinetic energy (KE) = 1/2 mv2 ….(5)
Substitute (4) in (5),
KE = 1/2 m(2as) = mas …(6)
Now, substituting equation (3) into (6), we have KE = m * (qE/m) * s. Then we find that the kinetic energy a charged particle q gains by moving a distance s, in an electric field E is depicted as KE = qEs.
7) A sphere S1 of radius R1 encloses a charge Q. If there is another concentric sphere S2 of radius R2, then if R2 is greater than R1 and if no extra charge is present in the space between S1 & S2, calculate the ratio of the electric flux via S1 & S2.
Ans – Assume Gauss’s law yields an equation for electric flux via a surface enclosing charge q.
ϕ = q/ϵ0
Where ϵ0 is the permittivity of the medium & now the electric flux via sphere S1, is denoted as,
ΦS1 = Q/ϵ0 …(1)
Since there is no extra charge on both spheres, flux via sphere S2 is mentioned as,
ϕS2 = Q/ϵ0 …(2)
We can now determine the flux ratio via spheres S1 & S2.
ϕS1/ϕS2 = (Q/ϵ0) / (Q/ϵ0)
ϕS1/ϕS2 = 1/1
Thus, we see that the desired ratio will be 1:1.
8) The electrostatic force on a small sphere of charge is 0.4µc due to other small spheres of charge -0.8µC in air is 0.2N.
a) How far apart are the two spheres?
Ans – As you know,
The electrostatic force on the 1st sphere is F = 0.2N
The charge on the first sphere is q1 = 0.4µC = 0.4 * 10-6 C
The charge on the Second sphere is q2 = 0.8µC = 0.8 * 10-6 C
The electrostatic force between the 2 spheres can be described using Coulomb’s law as:
F = (q1q2) / 4πϵ0r2 …(1)
Where, ϵ0 = Permittivity of free space
and 1/4πϵ0 = 9 * 109 Nm2C2
Now rearrange the 1st equation, r2 = (q1q2) / 4πϵ0F
Substitute this occurred equation,
r2 = (0.4 * 10-6 * (- 0.8) * 10-6 * 9 * 109) / 0.2
r2 = 144 * 10-4
r = sqrt(144 * 10-4)
r = 0.12m.
We found the distance between the two spheres to be 0.12m.
b) Calculate the force on the 2nd sphere compared to the 1st one!
Ans – As the gravitational attraction by every sphere to another is equal, the force applied on the second sphere compared to the first will be 0.2N.
9) An electron with a velocity of 2.0 × 106 ms-1 which is moving along x-axis enters the region between the two charged plates. The separation between the parallel plates of length L is 0.5 cm & the electric field is 9.1 × 102 N/C. Find where on the upper plate the electron will hit. (|e| = 1.6 × 10-19 C, me = 9.1 × 10-31 kg)
Ans – The element velocity is given by Vx = 2.0 * 106 ms-1.
Since the distance between the 2 plates is d = 0.5 cm/0.005 m.
The electric field between the plates is denoted as E = 9.1 * 102 N/C.
The charge of an electron is given as e = 1.6 * 10-19 C.
The mass of an electron is me = 9.1 * 10-31 kg.
For instance, s will be the deflection during the electron hits the upper plate at the end of plate L. Then, we can calculate the deflection using the following formula:
Substituting with occurred values
L = 1.6 * 10-2 = 1.6cm
The electron will hit the top plate after covering a distance of 1.6cm.