Title 10-electrostatics-.pdf ✅

ELECTROSTATICS 1. (C) kq 5 = kq 4 + Vind . 2. (C) If another shell is kept upside down over it complete a sphere, net field should become zero. 3. (B) Net downwards acceleration on body of mass m = (g + qE m ) = anet If E = uniform electric field in downwards direction If it hits after time t = 2v anet = 2v [g+ qE m ] at maximum height vf = 0 planet vf 2 = vi 2 − 2anet h ⇒ In uniform field (Gravitation + Electric field time to reach highest point = t/2 ] v 2 = 2 [g + qE m ] h ΔV between ground and highest point : ΔV = (E)h anet = 2v t ⇒ (g + qE m ) = 2v t ⇒ q m E = ( 2v t − g) E = m q ( 2v t − g) and h = (average velocity) × t ⇒ h = v 2 × t So ΔV = m q ( 2v t − g) ( v 2 t) = mv q (v − gt 2 ) 4. (C) If q denotes the charge in the medium from R to r, then E = K(Q+q) r 2 q can be calculated by integrating the charge dq in a thin shell of radius r, thickness dr. This gives q = 2πα(r 2 − R 2 ) i.e, E = K(Q+2πα(r 2−R 2 )) r 2 Electric field will be uniform if coefficient of r is made zero ⇒ kQ r 2 = 2παk R 2 r 2 ⇒ Q = 2παR 2 5. (D) E = λ 2πε0x + λ 2πε0(3a−x) = − dV dx E = λ 2πε0x + λ 2πε0(3a − x) = − dV dx ∴ ∫ − dV = λ 2πε0 ∫ dx x + ∫ dx 3a − x = λ 2πε0 {[ln x]a 2a − [ln (3a − x)]a 2a} = λ 2πε0 [ln 2 + ln 2] = λ 2πε0 ⋅ 2ln 2 = VA − VB ∴ work = (VA − VB)q0 = λq0ln 2 πε0 6. (B) For earthed conductor [the inner shell], V = 0 Here V = 1 4πε0 [ q 3r + q ′ r ] where q ′ is charge that would appear on inner shell as it is grounded ⇒ q ′ = −q/3 Hence, the charge flown to earth = 0 − (−q/3) = q/3 7. (B) θ1 = θ2 Enet = E⊥ ∝ 1 l E11 = 0
8. (A) r = 2Rcos θ dV = k dq r dq = σ((2θ)r)dr ⇒ dV = k σ(2θ)2Rcos θ(−2Rsin θdθ) 2Rcos θ V = ∫ dV = −4kσR∫0 π/2 θsin θdθ = σR πε0 . 9. (B) The system can be treated as a system of two dipoles having dipole moment p = qa each. If you think a little, you will realize that the dipoles are perpendicular to each other. Obviously, the net dipole moment is qa√2. 10. (D) E(x) = Qx 4πε0 (R2+x 2) 3/2 ≃ Qx 4πε0R3 (for x << R ) F = Qq 4πε0R3 x (Towards mean position) ⇒ T = 2π√ 4πε0mR3 Qq 11. (B) Charge distribution is as shown : σ1 = σ2 = σ3 ⇒ Q1 4πR2 = Q1 + Q2 4π(2R) 2 = Q1 + Q2 + Q3 4π(3R) 2 On solving we get ; Q1:Q2:Q3 = 1: 3: 5 On solving we get 12. (C) 13. (A) Field at P due to outside changes = K(2Q) (2R) 2 − KQ (4R) 2 = 7KQ 16R2 towards O 14. (A) ( using V(r, θ) = pcos θ 4π ∈0 r 2 ) ⇒ net potential at ( r 2 , √3r 2 , 0) is pcos 30∘ 4πε0r 2 + pcos 60∘ 4πε0r 3 = √3p 2(4πε0r 2) + pcos 60∘ 4πε0r 2 = p(√3 + 1) 8πε0r 2 15. (B) f = Kq1q2 4π∈r 2 and f 3 = K(q1−q2 ) 2 4π∈0r 24 where q1 and q2 be initial charges. ⇒ 4q1q2 = 3q1 2 + 3q2 2 − 6q1q2; Let q1 q2 = x ⇒ 3x 2 − 10x + 3 = 0 ⇒ x = 3 or 1 3 16. (AC) If forces due to both the fields cancel out then (A). If forces due to both fields are along the same line and the particle is given the initial velocity
along the same line, then also (A). In any other case, the net force acting on the particle will be uniform. From basic 2-D motion, we know that if acceleration is uniform but not parallel to initial velocity, the trajectory of the particle is a parabola. 17. (AD) Net charge on the surface of conductor is zero. Hence, potential at the center, due to charges on the surface of conductor is zero (as the center is equidistant from all points on the surface). Hence (A). At point B, total potential is same as that at center since volume of a conductor is an equipotential volume. Subtracting the contribution of point charge q from the total potential will give the potential at B due to charges induced at the surface of sphere. 18. (ABD)For (A), use the fact that potential at center of sphere, due to Q2 will be exactly cancelled by charge −Q2 induced on the surface of the cavity. Hence (A). Potential at any point at surface of cavity is equal to potential at the centre of sphere. Also, electric field inside cavity is known. So, potential at any general point inside cavity can be found be performing the line integral of electric field from the surface of cavity upto that point. This will solve (B). Potential outside the sphere cannot be found easily as the charge on the outer surface will be non- uniform. Hence C is incorrect as the given answer for uniform Q2 + Q3. (D)can be calculated by making the net potential at the center of sphere zero and finding new charge on sphere. 19. (AB) x0 = deformation in equilibrium state. 2Kx0 = σ ε0 ⋅ q ∴ x0 = σq 2Kε0 Springs are connected in parallel Keq = 2 K Angular frequency = √ 2K m 20. (AD) We know that for constant r, electric field due to a dipole is maximum at axial point and minimum at equatorial point. Magnitude of electric field continuously decreases from 2KP/r 3 to KP/r 3 as we move from an axial point to an equatorial point, keeping r constant. In the present case, if E = 10 N/C at any point, then it cannot be less than 5 N/C or more than 20 N/C at any point. 21. (BCD)Even if the metallic sphere is neutral, it will still be attracted towards the positive charge due to induced charges. 22. (BCD) (A) is incorrect by conservation of charge. (B) is correct. The net field inside the sphere is zero only because the electric field due to induced charges cancels the external electric field at all points inside the sphere. (C) is a basic property of conductors (D) As stated in part (B), the electric field lines are eliminated within the spherical region. 23. (BC) E = kq a2 ⇒ Enet = 3Ecos θ = 3 × kq a2 √2 √3 = √6kq a 24. (AC) Potential due to a uniformly charged spherical shell is given by: V = KQ r for r > R and V = KQ R for r < R 25. (AB) xa + xb + (Q2 + x)c = 0 ⇒ x = −Q2c a+b+c Q1 = y + x + y − Q2 − x ⇒ y = Q1+Q2 2 Potential different between A& B is V = Q2ca (a+b+c)Sε0 . Similarly, p.d between C&D depends upon Q2 26. (B) In part B, difference is due to the mass of electrons. Part (C) will not be true for a curved electric field line. 27. (ABD) Ex = 10 V/m, E ≥ Ex 28. (ABCD) (A) Net charge enclosed by Gaussian surface is zero. (B) Net charge enclosed by Gaussian surface is zero. (C) As the change is displaced towards sphere amount of negative charge induced on right half of
sphere will increase hence net flux through right hemispherical closed Gaussian surface increases. (D) Same reason as (C) and hence charge distribution on outer surface of sphere will change. 29. (AD) Uniform charge on outer surface therefore potential is q 4πε0R 30. (ABC) VA = VB ⇒ work = 0 31. (ABC) If moved slightly along x-direction, say towards left, the attractive force of left wire will be more than the attractive force of the right wire. Hence equilibrium will be unstable in this direction. If moved in y-direction, the force will remain zero. If moved in z-direction, electron will be attracted back towards the wires. 32. (ABCD) Charge on a1 = (r1θ)λ and Charge on a2 = (r2θ)λ Ratio of charges = r1 r2 E1, Field produced by a1 = K[(r1θ)λ] r1 2 = KQλ r1 ; E2, Field produced by a2 = KQλ r2 as r2 > r1 Therefore E1 > E2 i.e. Net field at A is towards a2. V1 = K. (r1θ)λ r1 = Kθλ, V1 = k(r2θ)λ r2 = kθλ ∴ V1 = V2 33. (B) At C and E, the electric fields will be subtracted and at D they will be added. 34. (AD) 35. (ACD) 36. (BC) Position A is position of equilibrium and B is critical position so tension and speed during motion is maximum at A and minimum at B. VMIN = VB = √geff l = √√2gl VMAX = VA = √5geff l = √5√2gl TMAX = TMIN = 6mggeff = 6√2mg 37. (AD) By conserving energy of the first ball, mg√⬚2mg 0 + q 2 4πε0 [ 1 r12 + 1 r13 + 1 r14 + ⋯ . . + 1 r1N ] = K1 + 0 And for the second ball, 0 + q 2 4πε0 [ 1 r23 + 1 r24 + ⋯ . + 1 r2N ] = K2 + 0 [It can be observed K1 > K2 ] ⇒ K = K1 − K2 = q 2 4πε0 ( 1 d ) ⇒ q = √4πε0dK 38. (ABCD)Flux through two circular surfaces is φ = 2 × q 2ε0 [1 − 4R √(3R) 2 + (4R) 2 ] = q 5ε0 ; φcurved = q ε0 − q 5ε0 = 4 5 q ε0 39. (AB) Consider a small element AB, θ is very small. Then AB = R(2θ) Change on AB is dQ = Q 2πR (2Rθ) = Qθ π 2Tsin θ = dQ ⋅ q 4πε0R2 = Qqθ 4π 2ε0R2 2T = Qqθ 4πε0R2 or T = Qq 8π 2ε0R2