Ch02 Static Electricity

PHYSICS 2ND PAPER CHAPTER 2 Static Electricity Question 1 [S. B. 2017] Two small spherical conductors 𝐴and 𝐵placed in air carry charges of +9 C and +16 C respectively. The distance between the spheres is 0.28 m. A point 𝐶is chosen on the line segment connecting the centres of 𝐴and 𝐵, at a distance of 0.12 m from 𝐴. 0.12 m 0.28 m 𝐴 𝐶 𝐵 +9 C +16 C (a) What is the electrostatic force exerted by sphere 𝐵on sphere 𝐴? (b) If a test charge of +1 C is placed at point 𝐶of the stem, will it experience any net electrostatic force? Give your opinion with mathematical reasoning. Solution (a) Here, Charge on sphere 𝐴 𝑞𝐴= +9 C Charge on sphere 𝐵 𝑞𝐵= +16 C Distance between spheres 𝑟= 0.28 m Coulomb constant 1 4𝜋𝜖0 = 9 × 109 N m2 C−2 According to Coulomb’s law, the magnitude of the electrostatic force exerted by sphere 𝐵on sphere 𝐴is: 𝐹𝐵𝐴= 1 4𝜋𝜖0 𝑞𝐴𝑞𝐵 𝑟2 = 9 × 109 × 9 × 16 (0.28)2 = 1.296 × 1012 0.0784 ≈1.653 × 1013 N Since both charges are positive, the force is mutually repulsive, acting along the line ® 𝐵𝐴(away from 𝐵). 𝐹𝐵𝐴= 1.65 × 1013 N (b) Here, ahbab.dev 1

Test charge at 𝐶 𝑞𝐶= +1 C Distance from 𝐴to 𝐶 𝑟𝐴𝐶= 0.12 m Distance from 𝐵to 𝐶 𝑟𝐵𝐶= 0.28 −0.12 = 0.16 m Electrostatic force on 𝑞𝐶due to charge 𝑞𝐴(repulsive, directed towards the right along ® 𝐴𝐵): 𝐹𝐴= 1 4𝜋𝜖0 𝑞𝐴𝑞𝐶 𝑟2 𝐴𝐶 = 9 × 109 × 9 × 1 (0.12)2 = 8.1 × 1010 0.0144 = 5.625 × 1012 N Electrostatic force on 𝑞𝐶due to charge 𝑞𝐵(repulsive, directed towards the left along ® 𝐵𝐴): 𝐹𝐵= 1 4𝜋𝜖0 𝑞𝐵𝑞𝐶 𝑟2 𝐵𝐶 = 9 × 109 × 16 × 1 (0.16)2 = 1.44 × 1011 0.0256 = 5.625 × 1012 N The net electrostatic force acting on the +1 C charge at point 𝐶is: 𝐹net = 𝐹𝐴−𝐹𝐵= 5.625 × 1012 −5.625 × 1012 = 0 N 𝐹net = 0 N Conclusion: Because the two forces are equal in magnitude and directed opposite to each other, the charge at point 𝐶will experience no resultant force and will remain in electrostatic equilibrium. ahbab.dev 2

Question 2 [D. B. 2017] In Figure (i), two static point charges are situated in air at points 𝑊and 𝑋, where 𝑞𝑊= +2 µC and 𝑞𝑋= +4 µC, separated by a distance of 2 mm. In Figure (ii), keeping the +2 µC charge at point 𝑊, the +4 µC charge is displaced from point 𝑋to point 𝑌, where 𝑋𝑌= 3 mm and ∠𝑊𝑋𝑌= 90◦. 𝑊 𝑋 +2 µC +4 µC 2 mm Figure (i) 90◦ 𝑊 𝑋 𝑌 +2 µC +4 µC 2 mm 3 mm Figure (ii) (a) Find the electrostatic force acting on the +2 µC charge in Figure (i). (b) Determine whether there will be any change in the electric potential at the position of the +4 µC charge between the configurations of Figure (i) and Figure (ii). Analyse mathematically. Solution (a) Here, Charge at 𝑊 𝑞𝑊= +2 µC = 2 × 10−6 C Charge at 𝑋 𝑞𝑋= +4 µC = 4 × 10−6 C Distance in Fig. (i) 𝑟1 = 2 mm = 2 × 10−3 m Coulomb constant 1 4𝜋𝜖0 = 9 × 109 N m2 C−2 Electrostatic repulsive force: 𝐹= 1 4𝜋𝜖0 𝑞𝑊𝑞𝑋 𝑟2 1 = 9 × 109 × (2 × 10−6) × (4 × 10−6) (2 × 10−3)2 = 9 × 109 × 8 × 10−12 4 × 10−6 = 1.80 × 104 N 𝐹= 1.80 × 104 N (b) The electric potential experienced by the +4 µC charge is created by the fixed source charge 𝑞𝑊= +2 µC at point 𝑊. In Figure (i), the distance of point 𝑋from 𝑊is 𝑟1 = 2 × 10−3 m: 𝑉1 = 1 4𝜋𝜖0 𝑞𝑊 𝑟1 = 9 × 109 × 2 × 1