Electromagnetism - Top 500 JEE Main Questions PDF

Electrostatics Top 500 Question Bank for JEE Main Physics...

Electrostatics Top 500 Question Bank for JEE Main Physics MathonGo Q1. Charge Q, is divided into two parts which are then kept some which direction(s) will +q be stable with respect to the displacement? distance apart. The force between them will be maximum if the two parts are having the charge - (1) Q/2 each (2) Q/4 and 3Q/4 (1) A and A 1 2 (3) Q/3 and 2Q/3 (2) B and B 1 2 (4) e and (Q − e), where e = electron charge (3) Both directions Q2. The diagram shows the arrangement of three small uniformly (4) Not stable charged spheres A, B and C. The arrows indicate the direction of the Q5. An α - particle and a proton are subjected to the same electric field, electrostatic forces acting between the spheres(for example, the left then the ratio of the forces acting on them is arrow on sphere A indicates the electrostatic force on sphere A due to (1) 2 : 3 sphere B ). At least two of the spheres are positively charged. Which (2) 1 : 2 sphere, if any, could be negatively charged? (3) 3 : 2 (4) 2 : 1 Q6. The path of a positively charged particle 1 through a rectangular region of a uniform electric field is as shown in the figure. What are the directions of the electric fields and the directions of deflection of particles 2, 3 and 4 ? (1) Sphere A (2) Sphere B (3) Sphere C (4) No sphere (1) Top; down; top; down Q3. The force of attraction between +5μC and −5μC charges kept at (2) Top; down; down; top some distance apart is 9 N. When the two charges are made to contact (3) Down; top; top; down and then separated again by the same distance the force acting between (4) Down; top; down; down them becomes (1) Infinite Q7. Two point charges of 1μC and −1μC are separated by a distance of (2) 9 × 10 9 N 100 Å. A point P is at a distance of 10 cm from the mid-point and on the (3) 1 N perpendicular bisector of the line joining the two charges. The electric (4) Zero field at P will be (1) 9 NC −1 Q4. Two identical point charges of charge −q are fixed as shown in the (2) 0.9 V m −1 figure below. A third charge +q is placed midway between the two (3) 90 V m −1 charges at the point P. Suppose this charge +q is displaced a small (4) 0.09 NC −1 distance from the point P in the directions indicated by the arrows, in Q8. Figure shows electric field lines in which an electric dipole p is placed as shown. Which of the following statements is If you want to solve these questions online, download the MARKS App from Google Play or visit https://web.getmarks.app Electrostatics Top 500 Question Bank for JEE Main Physics MathonGo correct? shown in figure ? (1) The dipole will not experience any force. (2) The dipole will experience a force towards right. (3) The dipole will experience a force towards left. (4) The dipole will experience a force upwards. Q9. Ten charges are placed on the circumference of a circle of radius R (1) λ ln 2 2πε0 with constant angular separation between successive charges. Alternate (2) − λ ln 2 2πε0 charges 1, 3, 5, 7, 9 have charge (+q) each, while 2, 4, 6, 8, 10 have (3) λ ln 2 4πε0 charge (−q) each. The potential V and the electric field E at the centre (4) − λ ln 2 4πε0 of the circle are respectively : (Take V = 0 at infinity) 10q Q13. Two charges +q and −q are kept apart. Then at any point on the (1) V = 4πϵ0 R ;E = 0 10q right bisector of line joining the two charges: (2) V = 0, E = 2 4πϵ0 R (1) The electric field strength is zero (3) V = 0; E = 0 (2) The electric potential is zero 10q 10q (4) V = ;E = (3) Both electric potential and electric field strength are zero. 4πϵ0 R 2 4πϵ0 R Q10. Positive and negative point charges of equal magnitude are kept at (4) Both electric potential and electric field strength are non-zero. and (0, 0, , respectively. The work done by the electric a −a (0, 0, ) ) → 2 2 Q14. A Point dipole p = −p x̂ 0 is kept at the origin. The potential and field when another positive point charge is moved from (−a, 0, 0) to electric field due to this dipole on the y− axis at a distance d are, (0, a, 0) is respectively: (Taken V = 0 at infinity) (1) positive → | p | → − p (1) 2 , 3 (2) negative 4πε0 d → 4πε0 d | p | (2) 0, (3) zero → 4πε0 d 3 → | p | p (4) depends on the path connecting the initial and final positions. (3) 3 , 3 4πε0 d 4πε0 d → − p (4) 0, Q11. Electric field lines 3 4πε0 d from positive charge and Q15. In Millikan's oil drop experiment an oil drop carrying a charge Q is in negative charge held stationary by a potential difference 2400 V between the plates. To (1) start; start keep a drop of half the radius stationary the potential difference had to be (2) start; end made 600 V. What is the charge on the second drop? (3) end; end Q (1) 4 (4) end; start (2) Q 2 (3) Q Q12. What will be the potential difference between two points A and B 3Q (4) in front of an infinitely long line of charge of linear charge density λ, as 2 Q16. A solid spherical conductor is given a charge. The electrostatic potential of the conductor is (1) Largest at the centre If you want to solve these questions online, download the MARKS App from Google Play or visit https://web.getmarks.app Electrostatics Top 500 Question Bank for JEE Main Physics MathonGo (2) Largest on the surface (4) 5 × 10 −6 J (3) Largest somewhere between the centre and the surface Q22. The electric flux for Gaussian surface A that encloses the charged (4) Constant throughout the conductor particles in free space. (Given, Q17. An electric dipole has the magnitude of its charge as q and its q1 = −14nC, q2 = 78.85nC, q3 = −56nC. ) dipole moment is p. It is placed in a uniform electric field E. If its dipole (1) 10 3 N m C 2 −1 moment is along the direction of the field, the force on it and its potential (2) 10 3 CN −1 m −2 energy are respectively (3) 6.32 × 10 3 N m C 2 −1 (1) 2q. E and minimum (4) 6.32 × 10 3 CN −1 m −2 (2) q.E and p.E (3) Zero and minimum Q23. In the figure shown here, A is a conducting sphere and B is a (4) q.E and maximum closed spherical surface. If a - q charge is placed at C near A, then the electric flux through the closed surface is- Q18. The ratio of electric field and potential (E/V) at mid-point of electric dipole of length l, for which separation is (1) 1/l (2) l (3) 2/l (4) None of these Q19. Two charges +3.2 × 10 −19 C and −3.2 × 10 −19 C are placed at 2.4 A apart from an electric dipole. It is placed in a uniform electric field (1) zero of intensity 4 × 10 5 Vm −1. The electric dipole moment is (2) positive (1) 15.36 × 10 −29 Cm (3) negative (2) 15.36 × 10 −19 Cm (4) can not be predicted (3) 7.68 × 10 −29 Cm (4) 7.68 × 10 −19 Cm Q24. A point charge causes an electric flux of −200 N m 2 C −1 to pass through spherical Gaussian surface of 10 cm radius Q20. The electric field intensity at a large distance x from an electric centered on the charge. If the radius of the Gaussian surface is doubled, dipole is proportional to the total electric flux passing through the surface is (1) 1 (1) −200Nm x 2 −1 C (2) 1 (2) −100Nm 2 −2 −1 x C (3) 1 (3) +200Nm 3 −1 x 2 C (4) 1 (4) −50Nm 4 x 2 −1 C Q21. The work done in moving a point charge of 10μC through a Q25. What is the nature of Gaussian surface involved in Gauss's law of distance of 3 cm along the equatorial axis of an electric dipole electrostatic? is (1) Scalar (1) zero (2) Electrical (2) 30 × 10 −6 J (3) Magnetic (3) 20 × 10 −6 J (4) Vector If you want to solve these questions online, download the MARKS App from Google Play or visit https://web.getmarks.app Electrostatics Top 500 Question Bank for JEE Main Physics MathonGo Answer Key Q1 (1) Q2 (1) Q3 (4) Q4 (2) Q5 (2) Q6 (1) Q7 (4) Q8 (3) Q9 (3) Q10 (3) Q11 (2) Q12 (1) Q13 (2) Q14 (4) Q15 (2) Q16 (4) Q17 (3) Q18 (4) Q19 (3) Q20 (3) Q21 (1) Q22 (1) Q23 (3) Q24 (1) Q25 (4) If you want to solve these questions online, download the MARKS App from Google Play or visit https://web.getmarks.app Electrostatics Top 500 Question Bank for JEE Main Physics MathonGo Q1. Let charges are q and (Q − q). kq(Q–q) force F = 2 = k 2 2 (qQ − q ) r r When the charge is at B and B , the net force acting on For maximum force 1 2 dF = 0 the charge will be towards the centre as shown. Therefore, dq k at B and B , there is stable equilibrium. When the charge 1 2 ∴ [Q − 2q] = 0 2 r Q is at A , the force of attraction from the charge on the left 1 ∴ q = 2 side will be more, so it moves towards left. When the ∴ Charge of other part charge is at A , the force of attraction from the charge on 2 = Q − q Q Q the right will be more, so it moves towards right. Hence, = Q − = 2 2 A1 and A are unstable. 2 Q2. Q5. F = Ee Q6. The positive charged particle moves as a parabolic path in electric field. Positive charge diverts in the direction of the field. ∴ The field must be towards the top. Particle 2 deflects down, 3 top. Particle 4 deflects down, 3 top. It can be seen from the diagram that only the sphere B and Q7. sphere C repel. Hence, they both must be of same type. According to the fact that at least two spheres are positively charged, therefore both spheres should be positively charged. Since attraction occurs for two remaining pairs it can be concluded that the sphere A is negatively charged. Q3. When we touch two oppositely charged bodies, charge redistribution takes place and both the bodies become equally charged because they are identical. Since both The point lies on the equatorial line of a short dipole. bodies were equally positive and equally negative initially, p 9×10 9 ( 10 −6 ×10 −8 ) ∴ E = = 3 3 after bringing them in contact with each other, they will 4πε0 r ( 10 −1 ) have zero charge which results in zero force between them. = 9 × 10 −2 = 0.09 N C −1. Q4. Q8. If you want to solve these questions online, download the MARKS App from Google Play or visit https://web.getmarks.app Electrostatics Top 500 Question Bank for JEE Main Physics MathonGo It is clear from the figure, that the dipole is present along the Z-axis and another point positive charge is moved on X- Y plane from A to B, both are axial points of the dipole. So, VA = VB = 0 Here, the space between the electric field lines is ⇒ W = QΔV = +1(VA − VB )= 0 increasing from left to right, so the strength of electric field decreases with the increase in the space between electric Hence, work done is zero. field lines. As a result force on charges also decreases from Q11. left to right. Let the electric field at the points of +q and − q charges Electric field lines starts from positive charge and end be E 1 and E2 respectively. From above discussion, at negative charge. E2 > E1. Generally it is considered that an electric field is described → → Force on the point charges is given as F = qE by the closeness of electric field lines. Since E 2 > E1 ⇒ F2 > F1 Since the electrostatic field is always directed away from Hence, the net force on the dipole is left side. positive charge and end towards negative charges, so the Q9. field lines must go away from positive charges to negative ones. Electric field lines never form closed loop. A → → Q12. V A − VB = − ∫ E. dr B r 2λ VA − V B = − ∫ dr 4π ε0 r 2r 2r 2λ 1 VA − V B = ∫ dr 4π ε0 r r r 2λ 2r = ln 4π ε0 r 2λ = ln 2 4π ε0 Q13. kq 5k(−q) Vnet = 5( R )+( R ) At equatorial point,due to dipole electric field is 1 p ⇒ Vnet = 0, [Qnet = 0] Ee = 4πε0 r 3 Enet = 0 , by symmetry But electric potential due to dipole at a point on the equiorial line is (directed from +q to -q) and V = 0. Q10. e Q +q q V = ⇒ Ve = − = 0 4πε0 r 4πε0 4πε0 → → → p p Q14. E = − 3 = − 3 4πε0 r 4πε0 d V = 0 If you want to solve these questions online, download the MARKS App from Google Play or visit https://web.getmarks.app Electrostatics Top 500 Question Bank for JEE Main Physics MathonGo Q15. Q18. Let AB be a dipole of length l. Hence O be the midpoint of the dipole. Let radius of drop is r. According to Millikan, for balance of drop QE = mg , where Q is the charge on the drop, E is electric field, m is the mass of the drop and g is the acceleration due to gravity. or Q V d =[( 4 3 πr )ρ]g 3 , where E = V d , Then the electric field at point O, due to +q charge 1 q E1 = 2 4πε0 l ( ) 2 4q 1 ⟹ E1 = (along OA) 4πε0 l 2 and the electric field at point O, due to -q charge 1 q E2 = 4πε0 2 l ( ) 2 1 4q ⟹ E2 = (along OA) where, V is potential difference and ρ is charge density of 4πε0 l 2 Hence, the total electric field at point O drop 3 E = E1 + E2 Q1 r1 V2 ∴ = ( ) × Q2 r2 V1 1 4q E = 2. 3 4πε0 2 l Q r 600 1 8q ∴ = ( r ) × = 2 ⟹ E = ….(i) Q2 2400 4πε0 l 2 2 Q Now, potential at O, due to +q charge ∴ Q2 = 2 1 2q ⟹ V1 = 4πε0 l Q16. For solid sphre charge is distributed uniformly so, Electric and potential at point O, due to – q charge potential is constant every where in the sphere ⟹ V2 = − 1 2q 4πε0 l kq ( equal to R , where k = 4πε0 1 ) Hence, total potential at point O, within or on the surface of conductor. V = 0......................(ii) Q17. On dividing Eq. (i) by Eq. (ii), we get E = ∞ V Q19. Electric dipole moment of a dipole is given by, p = qd (where, q is the charge and d is the distance The total force on dipole is zero because F = qE is applied on each charge but in between the charges) → → Given that q = ±3. 2 × 10 −19 C and opposite direction. The potential energy is U = − p ⋅ E , which is ∘ −10 → d = 2. 4 A = 2. 4 × 10 m → minimum when p and E are parallel. If you want to solve these questions online, download the MARKS App from Google Play or visit https://web.getmarks.app Electrostatics Top 500 Question Bank for JEE Main Physics MathonGo On substituting the given values in the above formula, we get, −19 −10 p =(3. 2 × 10 )(2. 4 × 10 ) −29 p = 7. 68 × 10 C m Q24. Q20. According to Gauss's theorem, electric flux through a In case of electric dipole, Electric field intensity in axial closed surface is given by, ∅ = ε0 q , 2p position is expressed as E = 4πϵ0 1 ⋅ 3 where, q is charge enclosed, ε is constant. 0 r 2p ∴ For distance x, E = 4πεo 1 ⋅ 3 ⇒ E ∝ 1 3 Here even if the radius is doubled, the charge enclosed is x x At a point far from a dipole in any direction, the magnitude the same, so the flux through the surface is also the same. of the electric field is proportional to the magnitude of Hence, flux is −200 Nm 2 C −1. the dipole moment and decreases with the cube of Q25. The Gauss's law in electrostatic given a relation between electric the distance. flux through any closed hypothetical surface (called a Gaussian surface) Q21. and the charge enclosed by the surface. The electric potential difference along the equatorial axis of an electric So, the nature is area vector. dipole is zero. The change of potential energy stored in charge 10 μC is equal to the work done on 10 μC to bring it from A to B. ∴ W = q Δ V = 0 Q22. From Gauss's Law we know that, qin ϕ = ε0 where, q is the net charge enclosed in So, for the given surface net charge enclosed can be calculated, The total charge enclosed in the surface A is, qin =(q1 + q2 + q3 )=(−14 + 78.85 − 56) nC −9 ⇒ qin = 8. 85 × 10 C Total flux through the surface A is, −9 qin 8.85×10 3 2 −1 ϕ = = = 10 N m C ε0 −12 8.85×10 Therefore electric flux for the gaussian surface encloses the charged particle is free space is 10 3 N m 2 C −1. Q23. Given that the figure shown here, A is a conducting sphere and B is a closed spherical surface. If a − q charge is placed at C near A Charge enclosed in Gaussian surface is – ve , so the electric flux (ϕ) = – ve If you want to solve these questions online, download the MARKS App from Google Play or visit https://web.getmarks.app Electrostatics Top 500 Question Bank for JEE Main Physics MathonGo If you want to solve these questions online, download the MARKS App from Google Play or visit https://web.getmarks.app Capacitance Top 500 Question Bank for JEE Main Physics MathonGo Q1. A circuit has a section AB, shown in the figure. The emf of the (4) 5C 12 source is equal to 10 V and the capacitance of capacitor are equal to Q4. For the arrangement of capacitors as shown in the circuit, the C1 = 1.0μF and C 2 = 2.0μF. If the potential difference between effective capacitance between point A and B is (capacitance points A and B, ϕ A − ϕB = 5.0 V , then the voltages V and V across 1 2 of each capacitor is 4μF ) each capacitors, C and C , respectively, are 1 2 (1) 4μF (1) V 1 = 5 3 V, V2 = 10 3 V (2) 2μF (2) V 1 = 10 3 V, V2 = 10 3 V (3) 1μF (3) V 1 = 10 3 V, V2 = 5 3 V (4) 8μF (4) none of these Q5. Initially charge on C is q and all other capacitors are uncharged. 1 0 Q2. In the given figure, three capacitors each of capacitance 6pF are If switch is closed at t = 0, then the charge on C after along time of 4 connected in series. The total capacitance of the combination becomes closing the switch is, (1) 2 × 10 −12 F (2) 3 × 10 −12 F (3) 6 × 10 −12 F (4) 9 × 10 −12 F Q3. Find the equivalent capacitance between A \& B if each capacitor (1) q has capacitance C ? 0 q0 (2) 2 q0 (3) 3 q0 (4) 4 Q6. A parallel plate capacitor has the space between its plates filled by two slabs of thickness d 2 each and dielectric constant K and K 1 2. d is the plate separation of the capacitor. The capacity of the capacitor is : 2ε0 d (1) K1 +K2 ( ) A K1 K2 2ε0 A (2) K1 K2 ( ) d K1 +K2 2ε0 A (3) (K1 + K2 ) (1) 4C A 3 (4) 2ε0 A K1 +K2 ( ) (2) 8C 3 d K1 K2 (3) 12C If you want to solve these questions online, download the MARKS App from Google Play or visit https://web.getmarks.app Capacitance Top 500 Question Bank for JEE Main Physics MathonGo Q7. A capacitor is half filled with dielectric (K = 2) as shown in plates. If the capacitance of the capacitor without copper slab is C and ′ diagram (1). If the same dielectric is to be filled in the same with copper slab is C , then ′ C is C capacitor as shown in diagram (2), then what would be the thickness of (1) √2 dielectric so that capacitor still has same (2) 2 capacitance ? (3) 1 (4) 1 √2 Q10. When a conducting slab fills the space between the two plates of a capacitor, its capacitance (1) becomes infinite. (2) becomes four times the original one. (3) remains same. (4) becomes zero. Q11. A parallel plate air capacitor has capacity C , distance of separation between plates is d and potential difference V is applied between the plates. The force of attraction between the plates of the capacitor is 2 (1) CV 2d 2 (2) CV (1) 2 d d 2 2 (3) 3 C V 2 (2) 3 d 2d 2 2 (4) 2 C V (3) 3 d 2d 4 (4) 4 d Q12. A parallel plate capacitor (without dielectric) is charged by a 3 battery and kept connected to the battery. A dielectric slab of dielectric Q8. Figure shows a parallel plate capacitor with three dielectric slabs of constant k is inserted between the plates, fully occupying the space same area A and thickness d. 3 between the plates. The energy density of electric field between the Find K eq =? plates will (1) increase k times. 2 (2) decrease k times. 2 (3) increase k times. (4) decrease k times. Q13. A parallel plate capacitor of capacitance 90pF is connected to a (1) 12 battery of emf 20 V. If a dielectric material of dielectric constant (2) 36 K = 5 3 is inserted between the plates, the magnitude of the induced (3) 18 charge will be: (4) 24 (1) 2.4nC (2) 0.9nC Q9. A slab of copper of thickness d is introduced between the plates of 2 (3) 1.2nC a parallel plate capacitor, where d is the separation between its two (4) 0.3nC If you want to solve these questions online, download the MARKS App from Google Play or visit https://web.getmarks.app Capacitance Top 500 Question Bank for JEE Main Physics MathonGo Q14. A parallel plate capacitor of 6μF is connected across 18 V battery Q15. A capacitor is charged and then made to discharge through a and charged. The battery is then disconnected and oil of dielectric resistance. The time constant is τ. In what time will the potential constant k = 2.1 is introduced between the plates. What will be the difference across the capacitor decrease by 10%? charge on capacitor? (1) τ ln 0.1 (1) 51.4μC (2) τ ln 0.9 (2) 108μC (3) τ ln 10 9 (3) 8.5μC (4) τ ln 11 10 (4) 92.5 × 10 2 C If you want to solve these questions online, download the MARKS App from Google Play or visit https://web.getmarks.app Capacitance Top 500 Question Bank for JEE Main Physics MathonGo Answer Key Q1 (3) Q2 (1) Q3 (2) Q4 (1) Q5 (3) Q6 (2) Q7 (1) Q8 (1) Q9 (2) Q10 (1) Q11 (1) Q12 (3) Q13 (3) Q14 (2) Q15 (3) If you want to solve these questions online, download the MARKS App from Google Play or visit https://web.getmarks.app Capacitance Top 500 Question Bank for JEE Main Physics MathonGo Q1. Let the potential at point A be V and at point B be V. A B From the question, we know that, VA − VB = 5 …(i). Let us write the voltage in the circuit starting from point Voltage drop across capacitor is given by V. q A. = C Therefore, q q ⇒ VA + C1 − 10 + C2 = VB. Substituting the values and using equation (i), q q −5 = + − 10 1 2 10 ⇒ q = μC 3 Hence, the voltage drop across C 1 , q V1 = C1 = 3×1 10 = 10 3 V. And voltage drop across C 2,. q 10 5 V2 = = = V C2 3×2 3 Q2. The reciprocal of equivalent capacitance of any number of capacitors joined in series is equal to sum of the reciprocals of individual capacitances. For n capacitors connected in series, total capacitance would be 1 i=n 1 = ∑ i=1 CS Ci Or, CS 1 = 1 C1 + 1 C2 + 1 C3 + … 1 1 1 1 3 1 ∴ = + + = = Cs 6 6 6 6 2 −12 ∴ CS = 2 pF = 2 × 10 F Q3. Take equipotential points and redraw circuit and identify series and parallel 2 C× C 4× 2 C 8 C Ceq. = 4 × = = 2 C+C 3 3 Q4. Given, capacitance of each capacitor, C = 4 μF The given circuit can be redrawn as shown in the figure: If you want to solve these questions online, download the MARKS App from Google Play or visit https://web.getmarks.app Capacitance Top 500 Question Bank for JEE Main Physics MathonGo ⇒ Ceq = 4 μF. Q5. In parallel, total charge distribution is the direct ratio of capacitance. Hence,. q0 q4 = 3 Clearly, this is a balanced Wheatstone bridge circuit. So, we can remove the middle capacitor. Q6. Let us understand this such that two capacitors are connected in series with K 1 and K2 as dielectric constant of the dielectric mediums present in between the plates. See the figure down below for better understanding. Now, the two 4 μF capacitors in each of upper and lower branches are connected in series. So, equivalent capacitance in each branch is 1 1 1 1 1 = + = + Ceq C1 C2 4 4 K1 ε0 A 2K1 ε0 A ∴ C' = 2 μF C1 = = d d ( ) 2 K2 ε0 A 2K2 ε0 A C2 = = d d ( ) 2 Now since both C 1 and C2 are arranged in series, we apply the formula for series combination of capacitor: 1 1 1 d d = + = + Cs C1 C2 2K1 ε0 A 2K2 ε0 A d K1 +K2 = ( ) 2ε0 A K1 K2 2ε0 A K1 K2 Cs = ( ) d K1 +K2 Now as the capacitors are in parallel, so equivalent capacitance is Ceq = C1 + C2 = 2 + 2 If you want to solve these questions online, download the MARKS App from Google Play or visit https://web.getmarks.app Capacitance Top 500 Question Bank for JEE Main Physics MathonGo Q7. Ceq = Keq = C 3 (K1 + K2 + K3 ) K1 +K2 +K3 36 Keq = = = 12 3 3 Q9. We know that the capacitance of parallel plate capacitor is Aε0 given by, C = d. ∈0 A C0 K C' = C" = 2d 2 ∴ C1 = C' + C' ' C0 C0 K C0 Figure shows the parallel plate capacitor with and without ∴ C1 = + = (1 + K) 2 2 2 copper slab. (Equivalent capacitance of 1 diagram) st And, the capacitance of parallel plate capacitor with metallic plate of thickness t is given by, Aε0 C ′ = d−t. Here, thickness, t = d 2. So, Aε0 2Aε0 C' = d = d. d− 2 Now, dividing C' by C , we get, C' C = 2 1 or 2. Q10. ∈0 A C2 = d – t+ t Capacitance is nothing but charge to potential difference K (Equivalent capacitance of 2 nd diagram) ratio. Capacitance (C) is given by Q Both are same hence : C = V C0 ∈0 A 2 (K + 1)= t where Q and V are charge and voltage, respectively. d – t+ K ∈0 A (2 + 1)= ∈0 A When a conducting slab is put in between, then potential t 2d d – t+ difference V between the plates of the capacitor becomes 2 3t 3d – = 2d 2 3t 2d zero. So, the capacitance becomes infinite. d = or t = 2 3 ∈0 A Q11. Attraction force between the plates, Q8. C = d 2 q ε0 A K 1 ∈0 A K1 C F = where q = CV and C = C1 = = 2Aε0 d 3d 3 2 2 2 C V CV K2 C ⇒ F = = 2Cd 2d C2 = 3 K3 C C3 = 3 Q12. If you want to solve these questions online, download the MARKS App from Google Play or visit https://web.getmarks.app Capacitance Top 500 Question Bank for JEE Main Physics MathonGo As the battery remains connected, potential difference ⇒ C' = kC = 5 3 C across the plate will remain the same. ⇒ C' = 5 3 × 90 × 10 −12 = 150 × 10 −12 F Here, C, U , V and C ′ ′ ,U ,V ′ are capacitance, energy Now the final charge Q f = C'V density of electric field and potential difference between ⇒ Qf = 150 × 10 −12 × 20 = 3000 × 10 −12 C the plates of the capacitor before insertion and after ⇒ Qf = 3 nC insertion of a slab. Induced charge, As the slab is introduced, Qinduced = Qf − Qi C ′ = (k)(C) Qinduced = 3 − 1. 8 = 1. 2 nC. 1 2 U = CV 2 Q14. ′ V = V U ′ = 1 ′ (C )(V ′ ) 2 When the capacitor is charged and then disconnected, 2 then form conservation of charge, the charge 'Q' across 2 ′ 1 U = (kC)(V ) 2 U ′ = k(U ) the plates remains constant, even when oil is introduced Energy density increases k times. between the plates. So the initial and final charge on the capacitor remains same, i.e., Q13. Qfinal = Qinitial = CV = 6 × 18 = 108 μC. Capacitance of parallel plate capacitor with air between t Q15. As we know the discharg