Title 1-electrostatics-.pdf ✅

Electrostatics 1. (A) tan θ = Fe mg = qE mg tan θ = qσ/2ε0 mg = σq 2ε0mg 2. (C) v = u + at, u = 0; v = qE m t; 1 2 mv 2 = q 2E 2 t 2 2m 3. (D) Since, the surface densities are equal, hence q1 4πr 2 = q2 4πR2 ( where (q1 + q2 = Q) Or q1 r 2 = q2 R2 = q1 + q2 r 2 + R2 = Q r 2 + R2 ∴ q1 = Q r 2 + R2 × r 2 and q2 = Q r 2 + R2 × R 2 So, potential at the common centre, V = q1 4πε0r + q2 4πε0R = 1 4πε0 ( q1 r + q2 R ) = 1 4πε0 [ Q R2 + r 2 × r 2 r + Q R2 + r 2 × R 2 R ] = 1 4πε0 Q(R + r) (R2 + r 2) 4. (CD) Only option C and D are true 5. (B) In air, F1 = 1 4πε0 q 2 r 2 ; In medium (k), F2 = 1 4πε0k q 2 r 2 ⇒ F1 F2 = k 1 6. (D) Let, the third charge −q is placed at a distance of ' x ' from the charge 2q. Then P.E. of the system is : U = k [ (2q)(8q) r − (2q)(q) x − (8q)(q) (r − x) ] U is minimum, when 2q 2 x + 8+q 2 r−x is maximum ; Solve for dy dx = 0 ⇒ x = r 3 , d 2y dx 2 = +ve i.e. at x = r 3 , y is minimum or U is maximum. So, there can't be any point between them where P.E. is minimum. 7. (C) When the two are joined by a metal wire, they become a single conductor. As charge can reside only on the outer surface of a conductor, the entire charge Q must flow to the outer sphere. 8. (D) Let an element of length dx, charge dq, at distance x from point O dV = k dq x Where, dq = Q L dx ∴ V = ∫L 2L kdq x = ∫L 2L k ( Q L ) dx x = Q 4πε0L ∫L 2L ( 1 x ) dx = Q 4πε0L [loge x]L 2L = Q 4πε0L [loge 2L − loge L] = Q 4πε0L [loge 2L L ] = Q 4πε0L ln (2) 9. (AB) Since V ≠ 0, there must be charges on the surface or inside itself. 10. (C) With each rotation, charge Q crosses any fixed point P near the ring. Number of rotations per second = ω/2π. ∴ Charge crossing P per second current Qω 2π
11. (C) W = (−Q)(ΔV) = −Q(Vf − Vi) = −Q(0 − Vi ),W = Q ⋅ ( 1 4πε0 ⋅ Q√2 a ) × 4 = √2Q 2 πε0a 12. (B) fnet = 2Fcos θ ∴ Fnet = − 2kq ( q 2 ) (√y 2 + a 2) 2 ⋅ y √y 2 + a 2 Fnet = − 2kq ( q 2 ) y (y 2 + a 2) 3/2 ⇒ −kq 2y a 3 ∝ −y [Negative sign indicate the net force is towards the mean position] 13. (A) E(x) = − dv dx , Ex = (2 − 2x), at x = 1E(x=1) = 0 14. (BC) E = − dv dr , since v is constant, E = 0 Since E = 0, there is no charge inside the region 15. (C) 16. (B) Force on −q1 F = 1 4πε0 q1q2 b 2 i + 1 4πε0 q1q3 a 2 [sin θi − cos θj] From above, x ′ component of force is : fx = q1 4πε0 ( q2 b 2 + q3 a 2 sin θ] Fx ∝ [ q2 b 2 + q3 a 2 sin θ] 17. (D) 18. (C) 19. (C) Complete the cube, adding five other faces. φcube = q ε0 ,φface = q 6ε0 20. (C) At P due to shell, potential V1 = q 4πε0R At P due to Q, potential V2 = Q 4πε0 R 2 = 2Q 4πε0R ∴ Net potential at P, V = V1 + V2 = q 4πε0R + 2Q 4πε0R 21. (BC) Work done an equipotential surface is zero also W = −∫A B E. dl 22. (B) VA = 1 4πε0 [ Q R + 2Q 2R ] = 1 4πε0 2Q R VB = 1 4πε0 3Q 2R , VA − VB = Q 8πε0R 23. (A) Using Gaussls theorem for radius r ∫ E⃗ . ds = 1 ε0 (Q + q) ⇒ E4πr 2 = 1 ε0 (Q + q) q = charge enclosed between x = a q = ∫a r A x 4πx 2dx = 4πA∫a r xdx = 4πA [ x 2 2 ] a r = 2πA(r 2 − a 2 ) q = charge enclosed between x = a and x = r.
Putting the value of q in equation (i), we get E × 4πr 2 = 1 ε0 [Q + 2πA(r 2 − a 2 )] E = 1 4πε0 [ Q r 2 + 2πA − 2πAa 2 r 2 ] E will be constant if it is independent of r ∴ Q r 2 = 2πAa 2 r 2 or A = Q 2πa2 24. (B) Consider a point in the emptied space whose position vector is r from the centre of the given sphere. Let the position vector of this point from the centre of cavity be r̅2. Let the position vector of the centre of cavity from the centre of the sphere be r 1. Clearly r 1 is constant. Consider the uniformly charged sphere without cavity. Electric field at r can be obtained using Gauss law. E13 × 4πr 2 = σ × 4πr 3 E1 = σ 3 r The direction of E⃗ 1 vector is along r , so we can write E1 = − σ 3 r̅2 Superposition of the two charges from the emptied space. To get the net electric field due to actual charge distribution superimpose the above two fields. E⃗ = E⃗ 1 + E⃗ 2 = σ 3 r 1This implies that the electric field in the emptied region is non-zero and uniform. 25. (C) We know E = − dV dr Here, ΔV and Δr are same for any pair of surfaces. So, E = constant Now, electric field inside the spherical charge distribution, E = ρ 3ε0 r ; E would be constant if ρr = constant ⇒ ρ(r) ∝ 1 r 26. (BD) According to gauss law option B and D are correct 27. (C) The electric field lines around the cylinder must resemble that due to a dipole. 28. (C) Electric field due to complete disc (R = 2a), E1 = σ 2ε0 [1 − x √R2 + x 2 ] = σ 2ε0 [1 − h √4a 2 + h 2 ] = σ 2ε0 [1 − h 2a ] (∵ h ≪< a) ⇒ ρ(r) ∝ 1 r 29. (D) Let a particle of charge q having velocity v approaches Q upto a closest distance r and if the velocity becomes 2v, the closest distance will be r ′ . The law of conservation of energy yields, kinetic energy of particle = electric potential energy between them at closest distance of approach. Or 1 2 mv 2 = 1 4πε0 Qq r Or 1 2 mv 2 = kQq r [∵ k = constant = 1 4πε0 ] and 1 2 m(2v) 2 = kQq r ′ 30. (C) On outer surface there will be no charge. So, Q2 = σ2 = 0 On inner surface total charge will be zero but charge distribution will be there so Q1 = 0 and σ1 ≠ 0 On inner surface total charge will be zero but charge distribution will be there so Q1 = 0 and σ1 ≠ 0
31. (AC) ACC. To Gauss law φ = qine ε0 = Q−2Q ε0 = −Q ε0 no contribution of charge 5Q in the flex through Gaussian surface as it is not enclosed by it. 32. (B) Electric flux φ = q ε0 where q = total charge enclosed by closed surface ∴ φ = 1.25 + 7 + 1 − 0.4 ε0 = 8.85C 8.885 × 10−12C 2 N−1 m−2 = 1012 N − m2 /C 33. (B) VA = Potential due to charge +q on ring A + Potential due to charge −q on ring B = 1 4πε0 ( q R − q d1 ) = 1 4πε0 ( q R − q √R2 + d 2 ) [∵ d1 = √R2 + d 2] Similarly, VB = 1 4πε0 (− q R + q √R2+d2 ) Potential difference, VA − VB = 1 4πε0 ( q R − q √R2 + d 2 ) − 1 4πε0 ( −q R + q √R2 + d 2 ) = 1 4πε0 q R + 1 4πε0 q R − 1 4πε0 q √R2 + d 2 − 1 4πε0 q √R2 + d 2 = q 2πε0 ( 1 R − 1 √R2 + d 2 ) 34. (ABCD) 35. (D) The flux passing through any enclosed area = Q ε0 For a Gaussian surface, ∮ E ⋅ ds = Q ε0 or E ⋅ 4πr 2 = Q ε0 or E = Q ε0 ⋅ πi 2 As flux is the total number of lines passing through the surface, for a given charge, it is always the charge enclosed Q/ε0 = φ. If area is doubled, the field will be half but the flux remains to same. 36. (C) In the given situation, it is a thick shell and not thin shell. So we have to consider line effect of induced charges. potential center vo = 1 4πεo [ q R − q 2r + q 3R ] = 1 4πεo ( 5q 6R ) 37. (A) 38. (D) 39. (A) 40. (-80) As we know, potential difference VA − V0 is : dV = −Edx ∫V0 VA dV = −∫0 2 30x 2dx VA − V0 = −30 × [ x 3 3 ] 0 2 = −10 × [2 3 − (0) 3 ] = −10 × 8 = −80 volt 41. (2) F = 1 4πε0 q(Q−q) r 2 For maximum repulsion force dF dq = Q − 2q = 0 Q q = 2 1 = 2 42. (0) 43. (0.25) Given E = q 4πε0x 2 Hence, the magnitude of the electric intensity at a distance 2x from charge q is E ′ = q 4πε0(2x) 2 = q 4πε0x 2 × 1 4 = E 4