O is the centre of the square.Ī charge of amount 1μC is placed at point O.įorce 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. The given figure shows a square of side 10 cm with four charges placed at its corners. What is the force on a charge of 1 μC placed at the centre of the square? Four point charges q A = 2 μC, q B = −5 μC, q C = 2 μC, and q D = −5 μC are located at the corners of a square ABCD of side 10 cm. A similar phenomenon is observed with many other pairs of bodies. This phenomenon is in consistence with the law of conservation of energy. When a glass rod is rubbed with a silk cloth, opposite natured charges appear on both the bodies. This is because equal amount of opposite charges annihilate each other. The net charge on the system of two rubbed bodies is zero. This phenomenon of charging is called charging by friction. Rubbing produces charges of equal magnitude but of opposite nature on the two bodies because charges are created in pairs. Explain how this observation is consistent with the law of conservation of charge. When a glass rod is rubbed with a silk cloth, charges appear on both. Therefore, it is ignored and it is considered that electric charge is continuous. Hence, quantization of electric charge is of no use on macroscopic scale. (b) In macroscopic or large scale charges, the charges used are huge as compared to the magnitude of electric charge. Hence, a body possesses total charge only in integral multiples of electric charge. This means that only integral (1, 2, …., n) number of electrons can be transferred from one body to the other. (a) Electric charge of a body is quantized. (b) Why can one ignore quantisation of electric charge when dealing with macroscopic i.e., large scale charges? (a) Explain the meaning of the statement ‘electric charge of a body is quantised’. This is the ratio of electric force to the gravitational force between a proton and an electron, keeping distance between them constant. Hence, the numerical value of the given ratio is M e and m p = Masses of electron and proton. Look up a Table of Physical Constants and determine the value of this ratio. Check that the ratio ke 2/G m em p is dimensionless. Therefore, the force on the second sphere due to the first is 0.2 N. (b) Both the spheres attract each other with the same force. The distance between the two spheres is 0.12 m. (a) Electrostatic force on the first sphere, F = 0.2 NĬharge on this sphere, q 1 = 0.4 μC = 0.4 × 10 −6 CĬharge on the second sphere, q 2 = − 0.8 μC = − 0.8 × 10 −6 CĮlectrostatic force between the spheres is given by the relation, (a) What is the distance between the two spheres? (b) What is the force on the second sphere due to the first? 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. Hence, the force between the given charged particles will be Since the nature of the charges is the same i.e. Putting the values in equation (1), we get, F12 is the force on charge q1 caused by charge q2.ĭistance between the spheres, r = 30 cm = 0.3 m Since, both the charges are positive, thus, the nature of force will be repulsive. The formula used to find the force, F is given as, 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? NCERT Solutions for class-12 Physics Chapter 1 Electric Charges and Fields is prepared by our senior and renowned teachers of Physics Wallah primary focus while solving these questions of class-12 in NCERT textbook, also do read theory of this Chapter 1 Electric Charges and Fields while going before solving the NCERT questions. You can download and share NCERT Solutions of Class 12 Physics from Physics Wallah. NCERT Solutions For Class 12 Physics chapter 1-Electric Charges And Fields
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