Tiêu đề 01. ELECTRIC CHARGES AND FIELDS.pdf ✅

01. ELECTRIC CHARGES AND FIELDS NEET PREPARATION (MEDIUM PHYSICS PAPER) Date: March 12, 2025 Dura on: 1:00:00 Total Marks: 180 INSTRUCTIONS INSTRUCTIONS PHYSICS 1. () : Explana on The dipole moments of the combina on is shown in the following figure The charge can be considered as a combi‐ na on of two charges each of charge . So forms dipole with . So we have two dipoles. Net dipole moment, i.e., 2. () : Explana on They will not experience any force if gravita‐ onal and electrosta c force are equal. 3. () : Explana on Let be each charge, According to Coloumb's law, the force of repul‐ sion between the two charges when kept in air is When half of the gap between the two charges is filled with the dielectric slab of dielectric con‐ stant , the new force of repulsion be‐ tween them becomes (where is the thickness of the slab) 4. () : Explana on Given, Radius of circular surface [Loop is parallel to - plane] Now, flux But −4q −2q +2q −2q pnet = √p 2 + p 2 + 2pp cos 60 ∘ = √3p = √3 (2ql) = 2√3ql(p = ql) ∣ ∣ ∣ −→ Fg ∣ ∣ ∣ = ∣ ∣ ∣ → Fe ∣ ∣ ∣ ⇒ G = ⋅ m2 (16×10 −2 ) 2 1 4πε0 q 2 (16×10 −2 ) 2 ⇒ = √4πε0G q m q F = 1 4πε0 q 2 r 2 (K = 16) F ′ = 1 4πε0 q 2 ((r−t)+t√K) 2 t = 1 4πε0 q 2 [(r− )+ √16] 2 r 2 r 2 (as t = , K = 16) r 2 = = 1 4πε0 q 2 ( +2r) 2 r 2 1 4πε0 q 2 ( r) 2 5 2 4F 25 E→ = E0 ^i + 2E0 ^j E0 = 100 N/C ⇒ E→ = 100^i + 200^j = 0.02 m Area = πr 2 = × 0.02 × 0.02 22 7 = 1.25 × 10 −3^im2 Y Z (φ) = EA cos θ φ = 0 ∘ φ = (100^i + 200^j) (1.25 × 10 −3^i) φ = 125 × 10 −3 Nm2/C = 0.125 Nm2/C
5. () : Explana on The field produced by an infinite sheet of charge density is . The resultant electric field is given by 6. () : Explana on Given, at from zero on -axis. 7. () : Explana on Though some enclosing surface around dipole can be taken. net charge But net enclosed charge/ So Gauss law can't be used So, correct op on is (2). 8. () : Explana on When (whether proton or electron ) en‐ ters perpendicular to it, then that par cle experiences force and both and follow curved (parabola) path. But, proton will move along and electron will move away from . So, correct op on is (2). 9. () : Explana on Given, Dipole moment, Torque ac ng on the dipole is , where is the angle between and . 10. () : Explana on Such metals, But this is due to So correct op on is (1). 11. () : Explana on Observing and are in opposite direc‐ ons, so angle between them is 12. () : Explana on 13. () : Explana on Number of atoms in given mass Transfer of electrons between balls Hence magnitude of charge gained by each ball. Force of a rac on between the balls 14. () : Explana on From gauss law This is the net flux coming out of the cube. Since a cube has 6 sides. So electric flux through any face is For two faces, flux σ σ 2ε0 E = E1 + E2 ⇒ E = + = σ1 2ε0 σ2 2ε0 σ1+σ2 2ε0 E1 = E2 x = 2 cm x E = Kq r 2 = ; = K ⋅ q1 (2) 2 K ⋅ q2 (x − 2) 2 10 4 40 (x − 2) 2 (x − 2) 2 = 16 ⇒ x − 2 = 4 ∴ x = 6 cm = 0 E→ = ( ε0) Q p e E→ p e E→ E→ q = 0.01C, l = 0.4 mm = 0.4 × 10 −3 m p = q(l) = 0.01 × 0.4 × 10 −3 = 4 × 10 −6C − m E = 10 dyne/C = 10 × 10 −5 N/C = 10 −4 N/C τ = pE sin θ θ →P E→ = 4 × 10 −6 × 10 −4 × sin 30 ∘ = 4 × 10 −10 × 1 2 = 2 × 10 −10N − m εr = ∞ F ′ = (F/εr) → 0 F ′ = 1 εr Enet p 180 ∘ Enet = E1 − E2 = − σ1 2ε0 σ2 2ε0 Enet = (σ1 −σ2) 2ε0 = × 6.02 × 10 23 = 9.48 × 10 10 22 63.5 = 9.48×10 22 10 6 = 9.48 × 10 16 q = 9.48 × 10 16 × 1.6 × 10 −19 = 0.015 C F = 9 × 10 9 × = 2 × 10 8N. (0.015) 2 (0.1) 2 φ = q ε0 φ1 = = q 6ε0 4πq q (4πε0) = 2φ1 = 4πq 3 (4πε0)
15. () : Explana on Electric field lines do not cross each other; they are con nuous curves. If they cross, it would mean 'two direc ons at one point to . But resultant is vector sums of fields by individual charges. Asser on is true but reason is false. So correct op on is (3). 16. () : Explana on The flux through the surface is 17. () : Explana on Inside a conductor in electrosta c equilibrium, the electric field is zero, including within cavi‐ es or voids within the conductor. The reason is also true, as it describes. The distribu on of charges in a conductor, which is primarily on its surface in the case of electrosta c equilibrium. Therefore, both the asser on and reason are true, and the reason correctly explains the as‐ ser on. So correct op on is (1). 18. () : Explana on 19. () : Explana on High energy X- rays cause photoelectrons from ball. Ball will become posi vely charged and de‐ flects in direc on of . So, correct op on is (3). 20. () : Explana on The given configura on of charges is combina‐ on of two dipoles and the point where the electric field required is on their equatorial line As the dipoles are perpendicular to each other 21. () : Explana on Using work energy theorem and equa ng the work done by the electric field and change in kine c energy, Hence 22. () : Explana on Tangent to electrosta c line gives direc on of and so line shall be curved. Also, they are con nuous. If there are breaks, it would mean there. So, correct op on is (4). 23. () : Explana on Electric field inside the conductor at point From Gauss law the charges on the inside faces of and are equal in magnitude and have opposite signs. From conserva on of charges on and As net electric field inside plate is zero. From 1,2, & 3 We get 24. () : Explana on The electric field ( ) at the centre of circular charged ring of radius is zero. 25. () : Explana on Two posi ve ions each carrying a charge are kept at a distance , then it is found that force of repulsion between them is where 26. () : Explana on Charge on any body is quan zed. 27. () : Explana on , so accelera on does not de‐ pend on velocity of the charge. Charge is invari‐ ant and it does not depend on frame Reason gives separate fact. So, correct op on is (2). E ̄ E ̄ φ = ExA = × 2 × 10 3 3 × 0.2 = 240 N. m2C−1 5 K. E = ΔU = qΔV = qEd = 4.2 × 10 −5 J ⇒ E→ E = (For each dipole) 1 4πε0 p x3 ENet = (p = √2aq) 1 4πε0 √2p x 3 = √2 = 1 4πεo q×√2a x 3 qa 2πεox 3 QE(x2 − x1) = (K2 − K1) ⇒ Q = (K2−K1) E(x2−x1) Q = = 8 × 10 −4C = 800μC (0.12−0) 300×(0.5−0) E→ E→ = 0 P = 0 A B A B q1 + q2 = Q1 (1) −q2 + q3 = Q2 (2) A = (3) q1 2Aε0 q2 2Aε0 −q2 = Q2−Q1 2 E R ∴ Force = qE = Zero q d = 1 4πε0 qq d 2 q = ne ∴ F = 4π 1 e ⇒ n = √ 0 n 2e 2 d 2 4πε0Fd 2 e 2 ∴ Q = ne n = = Q e 3.2 × 10 −6 1.6 × 10 −19 ⇒ n = 2 × 10 13 a = = E m qE m
28. () : Explana on Force on each sphere When both the spheres are allowed to touch each other, the charges get redistributed. Charge on each sphere Force on each sphere According to ques on, 29. () : Explana on Electric field of at the loca on of Electric field of at the loca on of 30. () : Explana on Let and be the two point charges. The force between the charges, at a separa on of . Suppose that force between the two charges become , when the charges are kept at a distance apart. Then or or 31. () : Explana on Flux does not depend on the size and shape of the close surface, and so, it remains same. 32. () : Explana on In metallic shield, due to , charge is induced At any point . Hollow shell is useful to block . So, correct op on is (1). 33. () : Explana on As the filed at the centre is zero for a uniformly closed ring, so the field at the centre will be equal and opposite to the field produced by the small element. 34. () : Explana on Imagine a hypothe cal cube with centre . Total flux through this cube is Flux through each face 35. () : Explana on Moving vehicle gets charged due to fric on, The inflammable material may catch fire due to the spark. If metallic rope is used, charge flows to ground. So correct op on is (1). 36. () : Explana on By Gauss law, we know that Here, Net electric flux, 37. () : Explana on As it is clear from figure As 38. () : Explana on From Coulomb's law, the force ac ng between two charges separated at a distance is given by Total force ac ng on charge is According to ques on, = = × 10 −18N kq1q1 2 −16k d2 = nC = −3nC (2−8) 2 = = × 10 −18N k(−3nC) 2 r 2 9k r 2 × 10 −18 = × 10 −18 ; r = d 16k d 2 9k r 2 3 4 −2Q Q1 E = K(−2Q)/r 2 Q −2Q E′ = K = − Q r 2 E 2 q1 q2 d F = ⋅ 1 4πε0 q1 q2 d2 F/3 ⋅ = 1 4πε0 q1 q2 x2 F 3 ⋅ = ( ⋅ ) 1 4πε0 q1 q2 x 2 1 3 1 4πε0 q1 q2 d 2 x = √3 d E→ ⇒ E→ = 0 E→ dq = (0.002π) Q 2πr = × = 2 × 10 −3C 1 2π(0.5) 2π 1000 E = = 7.2 × 10 7NC 9×10 −1 9×2×10 −3 (0.5) 2 O q εo = × = 1.88 × 10 5Nm2c 1 −1 6 q εo φ = q ε0 φ = φ2 − φ1 = 9 × 10 6 − 6 × 10 6 = ⇒ q = 3 × 10 6 × ε q ε0 ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ AB + ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ BC = ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ AC E = − = − = dV dr (VC−VA) d VA−VC d ∴ VA − VC = Ed (q1 , q2) r F = 1 4πε0 q1q2 r 2 Q F = + 1 4πε0 qQ (l/2) 2 1 4πε0 4Q ⋅ Q (l) 2 F = 0 ∴ + ⋅ = 0 ⇒ q = −Q 1 4πε0 qQ (l/2) 2 1 4πε0 4Q2 (l) 2