Question

1. Write Coulomb's Law. What are the units for each variable in the formula? 2. Two objects with charges of +1.0 C and 1.0 C are separated by 1.0km. Find the magnitude of the force that either charge exerts on the other. 3. Two charges attract each other with a force of 16N. What will be the new force if the distance between them is reduced to 1/2 of its original value? 4. Draw the electric field lines around a positive charge. Draw the electric field lines around a negative charge. 5. Which direction will a positive charge move if it is placed in the electric field surrounding a proton? Which direction will a positive charge move if it is placed in the electric field surrounding an electron? 6. What is the formula for the electric field? What is the Sl unit for the electric field?

Answer 1

Ans. I :- # Coulomb's luw!:- Statement: - "The force of attraction or sepulsion between two point charges cet mest is directy proportioncel to the product of the charges and inveysely proportional to the Changes and inversely proportioncel to the square of the distance between them. => Let 9..and 92 be two point charges at hest with respect to each other separated by the a distance 4. The muqnitude of the force between them is, 92 C FX9,92 sy 2 et whilst womerton F = 2,93 - white Unit if XNm² store open 42 Where F = Force between charges - N. (Newton) 9,= ¢ (coulomb) 92 = ¢ C coulombs 4 = m (meter) lecolomb's constunt. h=I → Eo = permittivity of the free. " LITTEO space o 60=8.85418541012 ime · Ketut 58.987557109 Nm2 4TTEO G 9x10° nm?

Ans:-2. Find: Fo? Given dates :- 9. = +1.04. 92=-1.0¢ 94 = .o km. - 1x 103 m. F=k9,92 42. = 9x109x1x6) clx103)2 =69x104) 106 --9x109-6 = -9xioºn = -9 KN

Ans. 3. Given duta - Fr= attraction. Force - 16 N 9 = rm a. 51 do ta case. 1 47 42 : 16 = kq,92 case.ii distance is reduce ka. Farkan gi bent ket .. Fe = 4 k29192 using eq.(1). in 29.(2), F2 = 4x16 F2 = 64 N

Ans. 4 raH(-→→→ Positive chceyge Electric field lines Negative charge Electric field lines Ans. 5. ws. → If positive charge put swirounding a proton, positive charge can move fell to proton (Septes repulsive force). Because Sume charge have repulsive force between them. 3. If positive charge put surrounding an. nequetive Celectron). Positive charge can move necey to electron. Cattraction force). Because different charge have attraction force betweer. them. Attraction fore Repulsive 0 favice opet

Ans:-6. The electric field is sometimes called field intensity ou erectori electric intensity. It is the strength of the electric field. → The electric intensity at any point in an electric field is the force exporte expe by ci unit positive test change kept at that point. , * The force acting on 9. is F. Then electric intensity cut that point is given by. E = F/ N where E-electseic field vendit . F- Force Yo-test charge & SI unit N/a and. mv.mt + For. single charge Cunit chwige) gabe EF E-k9,92 take 1-20, 22=2 a 7xlo. E-k 40% - kq