Unreasonable Results A wrecking yard inventor wants to pick up cars by charging a 0.400 m diameter ball and inducing an equal and opposite charge on the car. If a car has a 1000 kg mass and the ball is to be able to lift it from a distance of 1.00 m: (a) What minimum charge must be used? (b) What is the electric field near the surface of the ball? (c) Why are these results unreasonable? (d) Which premise or assumption is responsible?

Solution: In the question a wrecking yard inventor wants to pick up cars by charging a 0.400 m diameter ball and inducing an equal and opposite charge on the car.. Then next if a car has a 1000 kg mass and the ball is to be able to lift it from a distance of 1.00 m. First we want to know about W recking yard inventor, The wrecking yard is the location of a business in dismantling where wrecked or decommissioned vehicles are brought, their usable parts are sold for use in operating vehicles. While the unstable metal parts, known as s crap metal parts, are sold to metal-recycling companies. The above figure is the Wrecking yard car parts Step 1 (a) What minimum charge must be used We need to find the minimum charge of the car used, through the coulomb’s law of electrostatic force, The Electrostatic force between car and ball is, kq 2 F = r2 Where, q is charge on the ball same charge is also in car, r is the distance of car and ball, and k is constant. The Electrostatic attractive force must be equal to gravitational force acting on the car. So the,Gravitational force acting on car is F = mg G Here, m is the mass of the car, g is acceleration due to gravity. Therefore, the above two equation is F = F G kq 2 Substitute, m g for F andG, r2 for F, in the above equation is, 2 kq2 = mg r 2 q = r mg k Take the square root on both sides, we get r mg q = k Substitute, 1000 kg for m , (9.80 m/s ) for g , 1.0 m for r, and 9.00×10 Nm /C for k, in the above equation is, 2 q = (1.00 m) (1000 kg) (9.80m/s ) (9.00×10 Nm /C )2 2 = 1.043 × 10 C -3 Therefore, the minimum charge used is 1.043 × 10 C -3 Step 2 (b) What is the electric field near the surface of the ball Electric field near the surface of ball due to charge q is, kq E = r2 Where, r is the radius of ball, k is constant whose value is depending on permittivity of medium, q is charge on the ball same charge is also in car, -3 Substitute, 1 .043 × 10 C for q 0 equation is (9.00×10 Nm ./C )(1.043 ×10 E = (0.20 m)(0.20 m) E = 2.35 ×10 N/C 8 Therefore, the electric field near the surface is 2.35 ×10 N/C