Tiêu đề 02. ELECTROSTATIC POTENTIAL AND CAPACITANCE(H).pdf ✅

NEET REVISION 02. ELECTROSTATIC POTENTIAL AND CAPACITANCE(H) NEET REVISION Date: March 18, 2025 Dura on: 1:00:00 Total Marks: 180 INSTRUCTIONS INSTRUCTIONS PHYSICS 1. () : Explana on Let be radius of the big drop and be radius of each small drop. If be charge on each small drop, then charge on the big drop is The poten al of each small drop is and that of the big drop is Dividing eqn. (3) by eqn. (4), we get As (given) 2. () : Explana on because So, if now the charge is doubled and taken from in‐ finity then 3. () : Explana on There are total capacitors which are in se‐ ries. So mes R r As volume of the big drop = volume of eight small drops ∴ πR3 = 8 ( πr 3) 4 3 4 3 R = 2r (1) q Q = 8 q (2) V = (3) 1 4πε0 q r V ′ = (4) 1 4 πε0 Q R = = V V ′ 1 4πε0 q r 1 4πε0 Q R q Q R r = = (Using eqns. (1) and (2)) q 8q 2r r 1 4 V = V ′ 1 4 V ′ = 20 V ∴ V = (20 V ) = 5 V 1 4 W∞→p(ext) = −W∞→p(el) = 10mJ △KE = 0 ∴ Vp = = = 1 V W∞→p(ext) 10μC 10μJ 10μC 1 = or W∞→p(ext) = 20μJ ⇒ W∞→p(ele) = −20μJ W∞→p(ext) 20μC (n − 1) = + + ...(n − 1) 1 Ceq 1 C 1 C = ⇒ Ceq = 1 Ceq (n − 1) C C n − 1
NEET REVISION 4. () : Explana on Using Direc on of shall be from higher poten al sur‐ face towards lower poten al surface. Direc on of be perpendicular to the equipoten al surface i.e., at with -axis. So, op on (3) is correct. 5. () : Explana on We know Here, 6. () : Explana on Using Gauss’s law in differen al from 7. () : Explana on Without dielectric, For par al inser on of dielectric, On pu ng , we get 8. () : Explana on Work done 9. () : Explana on There is no poten al difference, hence no charge flows. Heat produce is zero. 10. () : Explana on The energy of the shell is equal to the work done by an external agent to bring the charge from to the surface. Let be the charge already formed on the surface. The small work done to bring from to the surface is 11. () : Explana on At closest distance , the whole kine c energy of charge is converted into poten al energy. In next case, dV = −E→ ⋅ → dr ⇒ ΔV = −EΔr cos θ ⇒ E = −ΔV Δr cos θ ⇒ E = −(20 − 10) 10 × 10 −2 cos 120 ∘ = −10 10 × 10 −2 (− sin 30 ∘) = 200 V /m E→ E (90 ∘ + 30 ∘) = 120 ∘ X E→ = − [ ^i + ^j + k^] ∂V ∂x ∂V ∂y ∂V ∂z = [2x + 3y − z] = 2 ∂V ∂x ∂ ∂x = [2x + 3y − z] = 3 ∂V ∂y ∂ ∂y = [2x + 3y − z] = −1 ∂V ∂z ∂ ∂z E = −(2 ^i + 3 ^j − k^) = −2 ^i − 3 ^j + k^ E = = 3ax −dV 2 dx = = 6ax or ρ(x) = 6aε0x ∂E ∂x ρ(x) ε0 C0 = ε0A d C = ε0A d − t (1 − ) 1 K t = d 3 = C C0 3k 2k + 1 = ∑All pairs 1 4πε0 qiqj rij = [ + 1 4πε0 (+q) × (−q) a (−q) × (+q) a + + ] + [ + ] = = (−4 + √2) (+q) × (−q) a (−q) × (−q) a 1 4πε0 (−q) × (−q) √2a (−q) × (+q) √2a (−4 + √2)q 2 4πε0a kq 2 a V2 initial = = 10 V 20 2 V5 initial = = 10 V 50 5 ∞ q dwext dq ∞ dWele = − (dU) = − (Uf − Ui) = − [ − ] qdq 4πε0R qdq 4πε0(∞) dWele = −qdq 4πε0R dWext = −dWele = qdq 4πε0R ⇒ Wert = ∫ Q 0 qdq 1 4πε0R Wext = Q2 8πε0R r q ∴ mv 2 = ⇒ r = ⋅ 1 2 1 4πε0 Q ⋅ q r 1 4πε0 2Q ⋅ q mv 2 r ′ = 1 4πε0 2Qq m(2v) 2 = r ′ = [ ⋅ ] 1 4 1 4πε0 2Qq mv 2 = r ′ = r 4
NEET REVISION 12. () : Explana on By argument of symmetry (it will be half of the po‐ ten al produced by the full sphere) 13. () : Explana on Consider a thin rectangular sheet of thickness The capacitance of the thin sheet is The equivalent capacitance is Where 14. () : Explana on Since there is no horizontal force Let and be the ini al and final speeds of the electron. Horizontal velocity will remain the same 15. () : Explana on 16. () : Explana on Before closing the switch Total charge on the two plates connected to point is zero. Similarly total charge on the two plates connected to point is zero. A er closing the switch 17. () : Explana on before A er inser ng the slab the capacitances are As capacitance remains same 18. () : Explana on Let be charge on each droplet. The poten al of each droplet is If be radius of the big drop formed, then Volume of the big drop Volume of 1000 droplets The charge on the drop is The poten al of the drop is As (given) ⇒ = 1 2 K(2Q) R KQ R ⇒ V = = = 300 V KQ R 9×10 9×5×10−9 15×10−2 dx dC = ε0(K0+λx)A dx = + +. . . = ∫ 1 C 1 dC1 1 dC2 1 dC = ∫ = d ∫ 0 1 C dx ε0A(k0+λx) 1 ε0A dx k0+λx = [ln (k0 + λx)] d 0 1 C 1 λε0A = In ( ) 1 λε0A k0+λd k0 C = ⇒ C = ε0Aλ ln( ) k0+λd k0 λdC0 ln(1+ ) λd k0 C0 = ε0A d u v ⇒ ⇒ u cos α = v cos β (1) = = ( ) 2 K.E.initial K.E.final mu 2 1 2 mv 2 1 2 cos β cos α U = CV 2 = ( ) V 1 2 2 1 2 ε0A x = [ ] dU dx ε0AV 2 2 −1 x2 dx dt = (x −2) dU dx −ε0AuV 2 2 a b C1 = ε0A 3 & Kε0A 3 ε0A 2.4 = ⇒ k = = 5 ε0A 3 . kε0A 3 ε0A 2.4 + kε0A 3 ε0A 2.4 3 0.6 q V = (1) 1 4πε0 q r R = πR 3 = 1000 ( πr 3) 4 3 4 3 R = 10r Q = 1000q V ′ = = 1 4πε0 Q R 1 4πε0 (1000q) 10r = 100 ( ) = 100V (using(1)) 1 4πε0 q r V = 1V ∴ V ′ = 100V = 100(1 V ) = 100 V
NEET REVISION 19. () : Explana on Energy stored where is small volume is constant. 20. () : Explana on The points having equal poten als are shown in the figure The circuit can be redrawn as The given circuit is a wheatstone bridge. 21. () : Explana on Put in equa on (1) 22. () : Explana on Conceptual Ques on 23. () : Explana on Electric poten al at point is given as As, Hence, 24. () : Explana on Poten al at the surface of a sphere 25. () : Explana on In isolated condi on the charge distribu on on each surface of capacitor and is The poten al difference across first capacitor is The poten al difference across second capacitor is Hence poten al difference across both the capaci‐ tors is same (hence ). The value of ca‐ pacitance in on connec ng both capacitors, charge does not flow from one capacitor to other. Net charge on upper and lower plates of is posi ve. Hence Asser on is true and Reason is false. So correct op on is (3). U = ∫ ε0E2dV 1 2 dV ∴ U = ε0E 2 ∫ dV 1 2 ∵ E ∴ U = ε0 ⋅ a 3 = 1 2 σ 2 4ε 2 0 σ 2a 3 8ε0 CPQ = 4μF Q1 + Q2 + 4 = 0 ⇒ (VP − 5) 1 + (VP − 3) C + 4 = 0 VP − 5 + VPC − 3C + 4 = 0 VP (1 + C) = 1 + 3C ⇒ VP = (1) 1+3C 1+C (VP − 1) 4 = 4 ⇒ ( − 1) = 1 1+3C 1+C 1 + 3C − 1 − C = 1 + C ⇒ C = 1μF (2) C ⇒ VP = 2V P V = 1 4πε0 qx cos θ r 2− cos 2θ x2 4 r >>> x V ∝ qx V = 1 4πε0 Q R Q = V .4πε0R = 9 × 10 5 × × 2 × 10 −2 = 2 × 10 −6C Wext = ΔU = q (VA − VC) = 1 × [ − ] = × [ − ] Wext = × 10 4 = 42 × 10 3 J = 42 kJ 1 9 × 10 9 1 4πε0 2 × 10 −6 25 × 10 −2 1 4πε0 2 × 10 −6 60 × 10 −2 1 4πε0 2 × 10 −6 10 −2 1 25 1 60 21 5 C1 C2 = q1 C1 1μF C1 = q2 C2 2μF C2 C2 = 2C1 ⇒ C1