Question

What is the magnitude and direction of the force on a proton that is placed at point B? (Note that the electron mentioned in problem #6b is removed from point A) Give your answer in the form "abc x 10^()* N. In the "Same/opposite direction as the electric field. Note that there is a comma followed by a space between the magnitude and the direction parts of the answer Answer: 11.00 x 10^(-17), same The proton is now removed from point B. Assume that the electric field at point is uniform and the space between the field lines at point C is 1/3 of what it is at point B. An electric dipole made up of two particles de and de separated by a distance of 16.8 nm, is now placed at point C. What is the magnitude of the torque on the dipole if the angle between the dipole moment and the electric field at C is 280 degrees? (Note Assume that the electric field is uniform over the dipole.) Give your answer in the form "abc X 10MLY) N. Answer A charge of 1256e is unifomiy distributed along a circular and of radius 28.2E c h subtenas an angle of 25 degrees see Figure 2) Determine the linear charge density along the larg Give your answer cuided to the

Answer 1

From answer 1), at point B, the force is given by

F = 1* 107N

We know the relation between electric field and Force is , F = qE, where q is the charge of the proton i.e, q = 1.6 x 10^(-19) C

---------------------(1)

110 1.6 10-19 F E = 0.625 102N/C 62.5N/C

The flux related to this field is, --------------------------(2)
where A is the area perpendicular to the field and E is the electric field.

=EA

Now at point C, Let the new flux be   --------------------------- (3) We assume area A to be same at both point B and C

= EA

Now its given that at point C, the space between the field lines is 1/3 of that at point B.

-----------------------(4)

= 3¢

Subsititutin equation (4) in (3) we get,

--------------------(5) this is the electric field strength at point C.

E 3E= 3 x 62.5 187.5N/C

Now the torque on the dipole is given by the relation,  

-----------------------------------------(6)

T = qdE' sin

here, q = 4e = 4 * 1.6*10^(-19) C= 6.4*10^(-19)C

d = separation between two charges = 16.8nm = 16.8 * 10(-9) m

E' = 187.5 N/C

Theta = 28 degree

16.8 10 187.5 sin(28°) 9455.04 x 10 28 6.4 10 T = 9.45 x 10 25 m