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'5'??? The Coulomb pael- In the previous lesson. we defined the value electric field strength, E, to be the ratio of the electric force, Fe. to the charge, q, experiencing the force. E: F! 9' If the electric field strength is known, then to calculate the electric force, Fe, acting on a charge, q, in an electric field, E, we use FB = qE The electric field, E, and the electric force, F8, are both vectors. When determining the electric force, both the magnitude (size), and the direction of the force should be stated. The electric field line pattern for a positive point charge and a negative point charge was discussed earlier. Atest charge of +5.00 C is placed a distance of2.00 m to the right of the point charge +q. The strength of the electric field produced by this charge is 10.0 N/C at the location of the charge q. The electric force on the charge q is Fe = CE = (+5.00C)(10.0N/C) = +50.0N The charge test charge, q, and the point charge producing the electric field, +q, are both positive, so that charges repel each other. The direction of the force is to the right (or east). In general, we can say that the electric field vector and the electric force vector are in the same direction if the charge brought into the field is positive. Atest charge of +5.00 C is placed a distance of2.00 m to the right of the point charge -q. The strength of the electric field produced by this charge is 10.0 N/C at the location of the charge q. The electric force on the charge q is Fe = CE = (+5.00C)(10.0N/C) = +50.0N The charge test charge, -q, and the point charge producing the electric field, +q, are opposite in charge, so that charges attract each other. The direction of the force is to the left (or west). In general, we can say that the electric field vector and the electric force vector are in opposite directions if the charge brought into the field is negative. Ml mw Electric Field and Force in One Dimension Duel: In our work in dynamics, we reviewed how to add vectors. If there exists at a place in space, electric fields caused by more than one source, then the fields will add together like vectors since electric field is a vector. In the situation below, there are two charges. The positive charge, (71, on the left creates an electric field, E,, of 10.0 N/C to the right at position P. The positive charge on the right, qz, creates and electric field, E2, of 15.0 NC to the left and position P. The total electric field at position Pis ET = E, + E2 =10.0 NO + (15.0 N/C) = 5.0 N/C or 5.0 MG west. If a charge of2.0 C is placed at position P, then the force on this charge is F = q'ET = (2.0 C)(5.0 N/C) = 10. N. The direction of this force is in the same direction as the electric field vector, that is west. Similarly, if a charge of 2.0 C is placed at P, then the force on this charge would be 10. N. But the direction of the force would be opposite to the direction of the electric field which is east. 337: Electric Field and Force in Two Dimensions DIXBII In the next situation, the two electric fields are at right angles relative to each other. The positive charge, (7,, creates and electric field of 10.0 N/C north while the charge, qz, creates and electric field of 15.0 N/C east. In adding together the two electric field vectors. We attach them head to tail. E; Er E2 E1 E2 The magnitude of the total electric field vector is [1" a}='l/(10.0NIC)2+(150NIC)J=18.0 NIC The direction of this vector is a = tan" M = 33.70 N ofE 15.0w 0 If a charge of2.0 C is placed at position P, then the force on this charge is F = qET = (2.0 C)(18.0 N/C) = 36. N. The direction of this force is in the same direction as the electric field vector, that is 33.70 N of E. Similarly, if a charge of 2.0 C is placed at P, then the force on this charge would be 36. N. But the direction of the force would be opposite to the direction of the electric field which is 337 8 of W. The physics of electric eld and force and point charges. A +4.0 C charge, +q1, is 1.0 m to the left of a 1.0 C charge, +q2. a. What is the magnitude and direction of the force exerted at the location of the +q2 charge? b. The charge +q2 is now removed. The +q1 charge produces an electric field at the location of point P. The magnitude of this electric field is 3.6 x 1010 N/C. Another charge +q3 to the right of point P produces an electric field of magnitude 1.6 X 1010 N/C at point P. What is the total electric field strength at point P and what force would a +5.0 C charge experience at this point? c. The charge (71 is now removed. It is replaced by a charge -q4. This charge is directly south of the point P and it produces an electric field of magnitude 2.6 x 1010 MC. Draw a diagram representing the two electric fields. Then determine the magnitude and direction of the total electric field