Title 01. ELECTRIC CHARGES AND FIELDS(H).pdf ✅

NEET REVISION 01. ELECTRIC CHARGES AND FIELDS(H) NEET REVISION Date: March 18, 2025 Dura on: 1:00:00 Total Marks: 180 INSTRUCTIONS INSTRUCTIONS PHYSICS 1. () : Explana on Let the spherical conductors and have same charge as . The electric force between them is where, being the distance between them. When third uncharged conductor is brought in contact with , then charge on each conductor When this conductor is now brought in contact with , then charge on each conductor Hence, electric force ac ng between and is 2. () : Explana on Electric field intensity at a point near and outside the surface of any charged conductor is given by (1) where, is charge density (charge per unit area). Electric field intensity due to uniformly charged in‐ finite thin plane sheet is given by (2) From eqs.(1) and (2) we have 3. () : Explana on Due to presence of test charge in front of posi‐ vely charged ball, there would be a redistribu on of charge on the ball. In the redistribu on of charge, there will be less charge on front half surface and more charge on the back half surface. As a result, the net force between ball and charge will decrease, the electric field is decreased. Thus, actual electric field will be greater than B C q F = 1 4πε0 q 2 r 2 r A B qA = qB = qA+qB 2 = = 0+q 2 q 2 A C qA = qC = qA+qC 2 = = (q/2)+q 2 3 q 4 B C F ′ = = 1 4πε0 qBqC r 2 1 4πε0 (q/2)(3q/4) r 2 = [ ] = 3 8 1 4πε0 q 2 r 2 3F 8 E1 = σ ε0 σ E2 = σ 2ε0 E1 = 2E2 q0 F i. e. F/q0
NEET REVISION 4. () : Explana on When and are in contact, Charge on Charge on ( Iden cal conduc ng bodies have half of total charge when touches each other) Again and are in contact, Charge on Since the distance between sphere and does not change, the new force between the spheres is, Using equa on (1), 5. () : Explana on The Fields produced by charges present in the cor‐ ners and get cancelled. Fields produced by other two opposite charges get added. The field at is 6. () : Explana on Gauss's law states that, the net electric flux through any closed surface is equal to the net charge inside the surface divided by i.e. Let electric flux linked with surfaces and are and , respec vely. i.e. Since, or Hence, (given, ) 7. () : Explana on From the concept of solid angle the flux of which passes through disc, It is given, or Hence, FPS = 16N = K ⋅ (1) Q2 r 2 P R P = = Q 2 R ∵ R S ∴ S = 3Q 4 P S (F ′ PS) F ′ PS = K = K ⋅ QP ⋅QS r 2 ⋅ Q 2 3Q 4 r 2 F ′ PS = = ⋅ K⋅Q×3Q 8×r 2 3 8 KQ2 r 2 F ′ PS = × 16 = 6 N (from(1)) 3 8 A C O Ecentre = E−2Q + E+2Q = = = k(2Q) (5a) 2 kQ (5a) 2 4kQ 25a2 ε0 φtotal = q ε0 A, B C φA, φB φC φtotal = φA + φB + φC φC = φA ∴ 2φA + φB = φtotal = q ε0 φA = ( − φB) 1 2 q ε0 φA = ( − φ) 1 2 q ε0 φB = φ Q φ = (1 − cos θ) Q 2ε0 φ = φtotal = 1 4 Q 4ε0 = (1 − cos θ) ⇒ cos θ = Q 4ε0 Q 2ε0 1 2 θ = 60 ∘ tan 60 ∘ = R b ⇒ R = b tan 60 ∘ ⇒ R = √3 × 5√3 = 15 m
NEET REVISION 8. () : Explana on Resolving the velocity of par cle parallel and per‐ pendicular to the plate. and Force on the charged par cle in the downward di‐ rec on normal is to the plate . Accelera on is , where is the mass of charged par ‐ cle. The par cle will not hit the upper plate, if the velocity component normal to plate becomes zero before reaching it, , , with , where , distance between the plates. Therefore, the maximum velocity for the par cle not to hit the upper plate (for this ) is 9. () : Explana on The total posi ve charge in the inner solid sphere is The total nega ve charge on the outer shell is .But both are equal in magnitude as the sum is zero. 10. () : Explana on From Gauss law, electric flux, As the sphere is of radius so it wil enclose only charge inside it (as charge is at distance which is outside the sphere). Hence, 11. () : Explana on Work is done by the electric field because 12. () : Explana on 13. () : Explana on Flux passing through Flux passing through So, ra o, 14. () : Explana on The two charges are shown in the figure. By considering the equilibrium of any charge From both the equa ons u∥ = u cos 45 ∘ = u √2 u⊥ = u sin 45 ∘ = u √2 eE a = eE/m m i. e. 0 = u⊥ 2 − 2ay y ≤ d d y = d = 2 cm u⊥ = √2 ay =  ⎷ 2 × 1.6 × 10 −19× 10 3 × 2 × 10 −2 1.6 × 10 −30 = 2 × 10 6 ms−1 umax = u⊥/ cos 45 0 = 2√2 × 10 6 ms −1 8 × 10 6 ms −1 R1 ∫ 0 4πr 2 dr. = + 4π ρ0 ρ0 r R2 1 2 −4πR 2 2σ = ∴ = √ R 2 1 R2 2 ρ0 2σ R2 R1 ρ0 2σ φ = qenclosed ε0 4a 5Q −2Q φ = 5Q ε0 θ = 0 ∘ , W = Fs cos θ → positive Fnet = 2F cos θ = 2 × = KQq (a2 + x2) x √a 2 + x 2 2KQqx (a2 + x2) 3/2 C1, φ1 = 2Q ε0 C2, φ2 = = 3Q + 2Q ε0 5Q ε0 φ1 : φ2 = = = 2 : 5 2Q/ε0 5Q/ε0 2 5 N cos θ = (1) q 2 4πε0d 2 N sin θ = mg (2) Tanθ = = mg4πε0d 2 q 2 √R2− d2 4 d/2 q = √ 2mgπε0d 3 √R2− d 2 4