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Unlock the NCERT Solutions for Class 12 Science Physics Chapter 2: Electrostatic Potential and Capacitance with clear, step-by-step explanations. These solutions have gained immense popularity among Physics students for their simplicity and effectiveness. Ideal for completing homework swiftly and gearing up for exams, the Electrostatic Potential and Capacitance Solutions ensure comprehensive understanding. Access all questions and answers from the NCERT Book of Class 12 Science Physics Chapter 2 without any cost. Page No 87: Question 2.1: Two charges 5 × 10−8 C and −3 × 10−8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero. ANSWER: There are two charges, Distance between the two charges, d = 16 cm = 0.16 m Consider a point P on the line joining the two charges, as shown in the given figure. r = Distance of point P from charge q1 Let the electric potential (V) at point P be zero. Potential at point P is the sum of potentials caused by charges q1 and q2 respectively. Where, = Permittivity of free space For V = 0, equation (i) reduces to Therefore, the potential is zero at a distance of 10 cm from the positive charge between the charges. Suppose point P is outside the system of two charges at a distance s from the negative charge, where potential is zero, as shown in the following figure. For this arrangement, potential is given by, For V = 0, equation (ii) reduces to Therefore, the potential is zero at a distance of 40 cm from the positive charge outside the system of charges. Page No 87: Question 2.2: A regular hexagon of side 10 cm has a charge 5 µC at each of its vertices. Calculate the potential at the centre of the hexagon. ANSWER: The given figure shows six equal amount of charges, q, at the vertices of a regular hexagon. Where, Charge, q = 5 µC = 5 × 10−6 C Side of the hexagon, l = AB = BC = CD = DE = EF = FA = 10 cm Distance of each vertex from centre O, d = 10 cm Electric potential at point O, Where, = Permittivity of free space Therefore, the potential at the centre of the hexagon is 2.7 × 106 V. Page No 87: Question 2.3: Two charges 2 μC and −2 µC are placed at points A and B 6 cm apart. (a) Identify an equipotential surface of the system. (b) What is the direction of the electric field at every point on this surface? ANSWER: (a) The situation is represented in the given figure. An equipotential surface is the plane on which total potential is zero everywhere. This plane is normal to line AB. The plane is located at the mid-point of line AB because the magnitude of charges is the same. (b) The direction of the electric field at every point on this surface is normal to the plane in the direction of AB. Page No 87: Question 2.4: A spherical conductor of radius 12 cm has a charge of 1.6 × 10−7C distributed uniformly on its surface. What is the electric field (a) Inside the sphere (b) Just outside the sphere (c) At a point 18 cm from the centre of the sphere? ANSWER: (a) Radius of the spherical conductor, r = 12 cm = 0.12 m Charge is uniformly distributed over the conductor, q = 1.6 × 10−7 C Electric field inside a spherical conductor is zero. This is because if there is field inside the conductor, then charges will move to neutralize it. (b) Electric field E just outside the conductor is given by the relation, Where, = Permittivity of free space Therefore, the electric field just outside the sphere is . (c) Electric field at a point 18 m from the centre of the sphere = E1 Distance of the point from the centre, d = 18 cm = 0.18 m Therefore, the electric field at a point 18 cm from the centre of the sphere is . Page No 87: Question 2.5: A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 10−12 F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6? ANSWER: Capacitance between the parallel plates of the capacitor, C = 8 pF Initially, distance between the parallel plates was d and it was filled with air. Dielectric constant of air, k = 1 Capacitance, C, is given by the formula, Where, A = Area of each plate = Permittivity of free space If distance between the plates is reduced to half, then new distance, d’ =  Dielectric constant of the substance filled in between the plates,  = 6 Hence, capacitance of the capacitor becomes Taking ratios of equations (i) and (ii), we obtain Therefore, the capacitance between the plates is 96 pF. Page No 87: Question 2.6: Three capacitors each of capacitance 9 pF are connected in series. (a) What is the total capacitance of the combination? (b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply? ANSWER: (a) Capacitance of each of the three capacitors, C = 9 pF Equivalent capacitance (C’) of the combination of the capacitors is given by the relation, Therefore, total capacitance of the combination is 3 pF3 pF. (b) Supply voltage, V = 120 V Potential difference (V‘) across each capacitor is equal to one-third of the supply voltage. Therefore, the potential difference across each capacitor is 40 V. Page No 87: Question 2.7: Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel. (a) What is the total capacitance of the combination? (b) Determine the charge on each capacitor if the combination is connected to a 100 V supply. ANSWER: (a) Capacitances of the given capacitors are For the parallel combination of the capacitors, equivalent capacitoris given by the algebraic sum, Therefore, total capacitance of the combination is 9 pF. (b) Supply voltage, V = 100 V The voltage through all the three capacitors is same = V = 100 V Charge on a capacitor of capacitance C and potential difference V is given by the relation, q = VC â€¦ (i) For C = 2 pF, …

NCERT Solutions for Class 12 Science Physics Chapter 2 – Electrostatic Potential And Capacitance Read More »

Explore the comprehensive NCERT Solutions for Class 12 Science Physics Chapter 1: Electric Charges and Fields, featuring straightforward, step-by-step explanations. These solutions have gained immense popularity among Class 12 Science students, proving invaluable for efficiently completing homework assignments and preparing for exams. The Physics Electric Charges and Fields Solutions offer a user-friendly resource to enhance your understanding of the subject. All questions and answers from Chapter 1 of the NCERT Book for Class 12 Science Physics are available here at no cost, ensuring easy access to valuable study material. Page No 46: Question 1.1: What is the force between two small charged spheres having charges of 2 × 10−7 C and 3 × 10−7 C placed 30 cm apart in air? ANSWER: Repulsive force of magnitude 6 × 10−3 N Charge on the first sphere, q1 = 2 × 10−7 C Charge on the second sphere, q2 = 3 × 10−7 C Distance between the spheres, r = 30 cm = 0.3 m Electrostatic force between the spheres is given by the relation, Where, ∈0 = Permittivity of free space Hence, force between the two small charged spheres is 6 × 10−3 N. The charges are of same nature. Hence, force between them will be repulsive. Page No 46: Question 1.2: The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge − 0.8 μC in air is 0.2 N. (a) What is the distance between the two spheres? (b) What is the force on the second sphere due to the first? ANSWER: (a) Electrostatic force on the first sphere, F = 0.2 N Charge on this sphere, q1 = 0.4 μC = 0.4 × 10−6 C Charge on the second sphere, q2= − 0.8 μC = − 0.8 × 10−6C Electrostatic force between the spheres is given by the relation, Where, ∈0 = Permittivity of free space The distance between the two spheres is 0.12 m. (b) Both the spheres attract each other with the same force. Therefore, the force on the second sphere due to the first is 0.2 N. Page No 46: Question 1.3: Check that the ratio ke2/G mempis dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify? ANSWER: The given ratio is . Where, G = Gravitational constant Its unit is N m2 kg−2. me and mp = Masses of electron and proton. Their unit is kg. e = Electric charge. Its unit is C. ∈0 = Permittivity of free space Its unit is N m2 C−2. Hence, the given ratio is dimensionless. e = 1.6 × 10−19 C G = 6.67 × 10−11 N m2 kg-2 me= 9.1 × 10−31 kg mp = 1.66 × 10−27 kg Hence, the numerical value of the given ratio is This is the ratio of electric force to the gravitational force between a proton and an electron, keeping distance between them constant. Page No 46: Question 1.4: (a) Explain the meaning of the statement ‘electric charge of a body is quantised’. (b) Why can one ignore quantisation of electric charge when dealing with macroscopic i.e., large scale charges? ANSWER: (a) Electric charge of a body is quantized. This means that only integral (1, 2, …., n) number of electrons can be transferred from one body to the other. Charges are not transferred in fraction. Hence, a body possesses total charge only in integral multiples of electric charge. (b) In macroscopic or large scale charges, the charges used are huge as compared to the magnitude of electric charge. Hence, quantization of electric charge is of no use on macroscopic scale. Therefore, it is ignored and it is considered that electric charge is continuous. Page No 46: Question 1.5: When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge. ANSWER: Rubbing produces charges of equal magnitude but of opposite nature on the two bodies because charges are created in pairs. This phenomenon of charging is called charging by friction. The net charge on the system of two rubbed bodies is zero. This is because equal amount of opposite charges annihilate each other. When a glass rod is rubbed with a silk cloth, opposite natured charges appear on both the bodies. This phenomenon is in consistence with the law of conservation of energy. A similar phenomenon is observed with many other pairs of bodies. Page No 46: Question 1.6: Four point charges qA = 2 μC, qB = −5 μC, qC = 2 μC, and qD = −5 μC are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1 μC placed at the centre of the square? ANSWER: The given figure shows a square of side 10 cm with four charges placed at its corners. O is the centre of the square. Where, (Sides) AB = BC = CD = AD = 10 cm (Diagonals) AC = BD = cm AO = OC = DO = OB = cm A charge of amount 1μC is placed at point O. Force of repulsion between charges placed at corner A and centre O is equal in magnitude but opposite in direction relative to the force of repulsion between the charges placed at corner C and centre O. Hence, they will cancel each other. Similarly, force of attraction between charges placed at corner B and centre O is equal in magnitude but opposite in direction relative to the force of attraction between the charges placed at corner D and centre O. Hence, they will also cancel each other. Therefore, net force caused by the four charges placed at the corner of the square on 1 μC charge at centre O is zero. Page No 46: Question 1.7: (a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not? (b) Explain why two field lines never cross each other at any point? ANSWER: (a) An electrostatic field line is a continuous curve because a charge experiences a continuous force when traced in an electrostatic field. The field line cannot have sudden breaks because the …

NCERT Solutions for Class 12 Science Physics Chapter 1 – Electric Charges And Fields Read More »