Title 02. ELECTROSTATIC POTENTIAL AND CAPACITANCE.pdf ✅

02. ELECTROSTATIC POTENTIAL AND CAPACITANCE NEET PREPARATION (MEDIUM PHYSICS PAPER) Date: March 12, 2025 Dura on: 1:00:00 Total Marks: 180 INSTRUCTIONS INSTRUCTIONS PHYSICS 1. () : Explana on Electric poten al is o en taken as zero at the Earth's surface, and this is related to the fact that the electric field due to the Earth's charges is negligible (very weak field exists) at its surface. So correct op on is (3). 2. () : Explana on The capacity of a conductor, which is its capacitance, depends on the size and shape of the conductor as well as the dielectric medium between the plates. So correct op on is (1). 3. () : Explana on For a parallel plate capacitor This corresponds to area of square of side which shows that one farad is very large unit of capacitance. 4. () : Explana on Let and be the radii of bigger and each smaller drop, resepc vely. The capacitance of a smaller spherical drop is The capacitance of bigger drop is [From Eq. (2)] 5. () : Explana on Poten al energy stored in dipole, and 6. () : Explana on Let be the poten al at the junc on . There is an independent loop in the circuit. The total charge on the independent loop is zero i.e., (Charge ) + (Charge on ) = (Charge on ) 7. () : Explana on The situa on is as shown in the figure. When they are connected by a wire the charge flows and both the spheres a ain the common poten al which is given by Since total charge is conserved Hence, from these two equa ons, Q ∝ V ⇒ Q = CV C = ∈0∈r A d C = ε0 A d ∴ A = = Cd ε0 1×10 −3 8.85×10 −12 = 1.13 × 10 8 m2 10.6 km R r ∴ πR3 = 8 × πr 4 3 3 4 3 ⇒ R = 2r (1) C = 4πε0r (2) C′ = 4πε0R = 2 × 4πε0r (∵ R = 2r) = 2C ∴ C = = μF (∵ C′ = 1μF) C′ 2 1 2 E = 2000 V /m, p = 10−29 cm, θ = 60 ∘ U = −→ p ⋅ → E = −pE cos θ = −10−29 × 2000 × cos 60 ∘ = −10−29 × 2000 × = −1 × 10−26 J 1 2 ⇒ N = −1 M = 26 N + M = −1 + 26 = 25 V O C1 C2 C3 q1 + q2 = (q1 + q2) C1 (V − V1) + C2 (V − V2) = C3 (V3 − V ) ⇒ V = = C1V1+C2V2+C3V3 C1+C2+C2 ΣViCi ΣCi V = = ⇒ = = Q1 4πε0R1 Q2 4πε0R2 Q1 Q2 R2 R1 0.02 0.01 ⇒ Q2 = 2Q1 Q1 + Q2 = 15 + 45 = 60 mC Q1 = 20 mC
8. () : Explana on Let be original charge of the capacitor and be capacitance of the capacitor. Ini al energy stored in the capacitor is When an addi onal charge of 2 is given to a capaci‐ tor, its charge on the capacitor be comes Now energy stored in the capacitor is As per ques on (Using (1) and (3)) (Using(2)) 9. () : Explana on The arrangement of three iden cal . capacitors to ob‐ tain dis nct effec ve capacitances is 4 , . (1) (2) (3) (4) 10. () : Explana on Maximum permissible field strength, Poten al on the surface of the sphere In case of charged spherical cell electric field ( ) and electric poten al ( ) at a point on its surface is given by and When is maximum, is minimum 11. () : Explana on Let be the charge on each drop, then From conserva on of volume Radius of bigger drop The poten al of the bigger drop is 12. () : Explana on The given situa on is equivalent to three capacitors and in series. Equivalent capacitance (3) is given by Here Also, 13. () : Explana on Capacitance of capacitor Rate of change of poten al 14. () : Explana on Op on (4) is correct 15. () : Explana on In a symmetrical conductor, the charges distribute themselves in such a way that the electric poten al is the same at all points on its surface, making it an equipoten al surface. The reason is valid because charges shall easily move within a conductor to make on surface So correct op on is (1). Q C C ∴ U = (1) Q2 2C C Q′ = Q + 2 (2) U ′ = (3) Q′2 2C U ′ = U + U = U 21 100 121 100 ∴ = ( ) Q ′2 2C 121 100 Q 2 2C Q′2 = Q2 or Q′ = Q 121 100 11 10 Q + 2 = Q 11 10 10(Q + 2) = 11Q 10Q + 20 = 11Q or Q = 20C i. e E = 3 × 10 6V m−1 = 1.5 × 10 6V E V E = K q/r 2 V = K q/r ⇒ r = V /E E r r = = 0.5 m = 50 cm 1.5×10 6V 3×10 6V /m Q V = 1 4πε0 Q r N πr 3 = πr 4 3 3 4 3 R = N 1/3r V ′ = = N 2/3V 1 4πε0 NQ N 1/3r C1, C2 C3 C1 = , C2 = , C3 = ε0K1A d1 ε0K2A d2 ε0K3A d3 = + + = ( + + ) 1 C 1 C1 1 C2 1 C3 1 ε0A d1 K1 d2 K2 d3 K3 A = 2m2 , K1 = 2, K2 = 3, K3 = 6 d1 = 0.4mm = 4 × 10−4m d2 = 0.6mm = 6 × 10−4m d3 = 1.2mm = 12 × 10−4m ε0 = 8.85 × 10−12C2N −1m−2 = = = 2 × 10−4m d1 K1 d2 K2 d3 K3 ∴ = ; C = 1 C 3 ε0A d1 K1 ε0AK1 3d1 = = 2.95 × 10−8F 8.85×10 −12×2 3×2×10 −4 C = 20μF ( ) = 3 V /s dV dt q = CV = C dq dt dV dt i = 20 × 10−6 × 3 = 60μA ΔV = 0