Question

Please help with #4 and 5 will rate thank you!

) wnat is the magnitude and sign of the charge a 190 lb human being would have to have in order for the electric force to just overcome her weight? (b) What is the magnitude of th people separated by 100 m and each possessing the charge calculated in part (a s the force attractive or repulsive? (Hint: approximate each individual as a particle.) 4. A 240 g cup of water (ignore the cup itself) is placed in a uniform electric fiel nat changes direction 150 x 109 (five billion) times. The electric field strength is E-10000 N/C. The d le moment of the water molecule is 6.2 x 10.30 C m and 1 g of water contains 3.35 x 1022 waterr ecules. Assume that the dipoles respond instantly to changes in the electric field and all kinetic en gy is deposited as heat. How much heat is deposited into the water? 5, A 5 g projectile with a charge of 30 μC is launched from rest to 300 m/s o r a distance of 0.5 m in a uniform static electric field. (a) What is the magnitude of the electric field?) What is the magnitude of the point charge that could produce this electric charge at a distance of 1 m from the charge? The following figure is for Questions 6-9. e ce between two 2 0 7 6. The electric field lines due to Qeft and qright are shown in the diagram at the left. What is the charge ee if nghel 1.5 nc? 7. Sketch the DIRECTION of the electric FIELD at point A 8. Sketch the DIRECTION of the electric FORCE on an electron at point A. ieft dnghr 1.5 nc 9. What is the magnitude of the electric FORCE on an electron at point A if the electric field magnitude at point A is E 30000 N/C? ch

Answer 1

4. An electric dipole is always aligned along the electric field.

work done in rotating the dipole through an angle  
An electric dipole is always aligned along the electric filed work done in rotating the dipole through an angle theta_1 to theta_2 is given by W = pE[cos theta_2 - cos theta_1). When the field changes its direction the dipole rotates from 0 to 180, hence work done W = 2 pE = 2*6. 2E - 30* 10000 = 12. 4E - 26J the field changes 150E + 9 times. Total work done on 1 dipole = 12. 4E - 26* 150E + 9 = 1. 86E - 12L number of water molecules in the cup = 240*3. 35E + 22 each water molecules acts as a dipole total work done by the field = 240*3. 35E + 22*1. 86E - 12 = 1. 5E + 13J total work done is released as heat heat generated = 1. 5E + 13J 5. Final vel v = 300 m/s distance = 0. 5 m v^2 = 2as: we assume initial vel = 0 acceleration a = v^2/2s = 300^2/2*0. 5 = 9. 0E + 4 m/s^2
₁ to
An electric dipole is always aligned along the electric filed work done in rotating the dipole through an angle theta_1 to theta_2 is given by W = pE[cos theta_2 - cos theta_1). When the field changes its direction the dipole rotates from 0 to 180, hence work done W = 2 pE = 2*6. 2E - 30* 10000 = 12. 4E - 26J the field changes 150E + 9 times. Total work done on 1 dipole = 12. 4E - 26* 150E + 9 = 1. 86E - 12L number of water molecules in the cup = 240*3. 35E + 22 each water molecules acts as a dipole total work done by the field = 240*3. 35E + 22*1. 86E - 12 = 1. 5E + 13J total work done is released as heat heat generated = 1. 5E + 13J 5. Final vel v = 300 m/s distance = 0. 5 m v^2 = 2as: we assume initial vel = 0 acceleration a = v^2/2s = 300^2/2*0. 5 = 9. 0E + 4 m/s^2
₂ is given by

W = pE(Cos
An electric dipole is always aligned along the electric filed work done in rotating the dipole through an angle theta_1 to theta_2 is given by W = pE[cos theta_2 - cos theta_1). When the field changes its direction the dipole rotates from 0 to 180, hence work done W = 2 pE = 2*6. 2E - 30* 10000 = 12. 4E - 26J the field changes 150E + 9 times. Total work done on 1 dipole = 12. 4E - 26* 150E + 9 = 1. 86E - 12L number of water molecules in the cup = 240*3. 35E + 22 each water molecules acts as a dipole total work done by the field = 240*3. 35E + 22*1. 86E - 12 = 1. 5E + 13J total work done is released as heat heat generated = 1. 5E + 13J 5. Final vel v = 300 m/s distance = 0. 5 m v^2 = 2as: we assume initial vel = 0 acceleration a = v^2/2s = 300^2/2*0. 5 = 9. 0E + 4 m/s^2
₂ - Cos
An electric dipole is always aligned along the electric filed work done in rotating the dipole through an angle theta_1 to theta_2 is given by W = pE[cos theta_2 - cos theta_1). When the field changes its direction the dipole rotates from 0 to 180, hence work done W = 2 pE = 2*6. 2E - 30* 10000 = 12. 4E - 26J the field changes 150E + 9 times. Total work done on 1 dipole = 12. 4E - 26* 150E + 9 = 1. 86E - 12L number of water molecules in the cup = 240*3. 35E + 22 each water molecules acts as a dipole total work done by the field = 240*3. 35E + 22*1. 86E - 12 = 1. 5E + 13J total work done is released as heat heat generated = 1. 5E + 13J 5. Final vel v = 300 m/s distance = 0. 5 m v^2 = 2as: we assume initial vel = 0 acceleration a = v^2/2s = 300^2/2*0. 5 = 9. 0E + 4 m/s^2
₁ )

when the field changes its direction the dipole rotates from 0 to 180 , hence work done

W = 2pE = 2*6.2E-30*10000 = 12.4 E-26 J

the field changes 150E+9 times

Total work done on 1 dipole = 12.4E-26*150E+9 = 1.86E-12 J

number of water molecules in the cup = 240*3.35E+22

each water molecules acts as a dipole

total work done by the field = 240*3.35E+22*1.86E-12 = 1.5 E+13 J

total work done is released as heat

heat generated = 1.5 E+13 J

5. final vel v = 300 m/s

distance s = 0.5 m

v² = 2as ; we assume initial vel =0

acceleration a = v²/2s   = 300²/2*0.5 = 9.0E+4 m/s²  

force on the particle F = ma = 5*E-3 * 9.0E+4 = 450 N

= Eq , electric force

electric field E = F/q = 450/30E-6 = 15E+6 N/C

if Q is the charge then

E = kQ/r² ; r = 1m

Q = E/k = 15 E+6 /9.0E+9 = 1.67 mC