Electric Force, Field & Potential A comparison

Example Three charges, q 1 = +8  C, q 2 = +6  C and q 3 = -4  C are arranged as shown below. Find the resultant force on the –4  C charge due to the others. Draw free-body diagram o -4  C q3q3 F1F1 F2F2 Note the directions of forces F 1 and F 2 on q 3 based on attraction/repulsion from q 1 and q cm 3 cm 5 cm 53.1 o +6  C -4  C +8  C q1q1 q2q2 q3q3 +

Example (Cont.) Next we find the forces F 1 and F 2 from Coulomb’s law. Take data from the figure and use SI units. F 1 = 115 N, 53.1 o S of W F 2 = 240 N, West Thus, we need to find resultant of two forces: +- 4 cm 3 cm 5 cm 53.1 o +6  C -4  C +8  C q1q1 q2q2 q3q3 F2F2 F1F1 +

Example 4 (Cont.) We find components of each force F 1 and F 2 (review vectors) o - -4  C q3q3 F 1 = 115 N F 1y F 1x F 1x = -(115 N) Cos 53.1 o = N F 1y = -(115 N) Sin 53.1 o = N Now look at force F 2 : F 2x = -240 N; F 2y = 0 R x =  F x ; R y =  F y R x = – 69.2 N – 240 N = -309 N R y = N – 0 = N F N R x = N R y = -240 N

Example 4 (Cont.) Next find resultant R from components F x and F y. (review vectors). R x = -309 N R y = N - -4  C q3q3 R y = N R x = -309 N  R We now find resultant R,  : R = 317 N Thus, the magnitude of the electric force is:

Example (Cont.) The resultant force is 317 N. We now need to determine the angle or direction of this force N N  R  N Or, the polar angle  is: Or, the polar angle  is:  = = The reference angle is: The reference angle is:   = S of W Resultant Force: R = 317 N,  =

Fields are vectors, too 1. Now, consider point P a distance r from +Q. 2. An electric field E exists at P if a test charge +q has a force F at that point. 3. The direction of the E is the same as the direction of a force on + (pos) charge. E 4. The magnitude of E is given by the formula: Electric Field Q. P r +q F +

The E-Field at a distance r from a single charge Q Q. r P Consider a test charge +q placed at P a distance r from Q. The outward force on +q is: The electric field E is therefore: ++q+q F Q. r P E

The Resultant Electric Field. The resultant field E in the vicinity of a number of point charges is equal to the vector sum of the fields due to each charge taken individually. Consider E for each charge. + -  q1q1 q2q2 q3q3 - A E1E1 E3E3 E2E2 ERER Vector Sum: E = E 1 + E 2 + E 3 Vector Sum: E = E 1 + E 2 + E 3 Directions are based on positive test charge. Magnitudes are from:

Electrical Potential, Work and Energy An external force F moves +q from A to B against the field force qE. Work = Fd = (qE)d At level B, the potential energy U is: U = qEd (Electrical) The E-field does negative work; External force does positive work. The external force F against the E-field increases the potential energy. If released the field gives work back. B A + +q d qE E FeFe

Work to Move a Charge Q  qE F  Work to move +q from A  to B.     rara rbrb At A: At B: Avg. Force: Distance: r a - r b

Electric Potential Potential Q. r Electric potential is another property of space allowing us to predict the P.E. of any charge q at a point. Electric Potential: joules per coulomb(J/C) The units are: joules per coulomb (J/C) For example, if the potential is 400 J/C at point P, a –2 nC charge at that point would have P.E. : U = qV = (-2 x C)(400 J/C); U = -800 nJ P

Calculating Electric Potential Potential Q. r P Electric Potential Energy and Potential: Substituting for U, we find V: The potential due to a positive charge is positive; The potential due to a negative charge is positive. (Use sign of charge.)

Potential For Multiple Charges The Electric Potential V in the vicinity of a number of charges is equal to the algebraic sum of the potentials due to each charge. + -  Q1Q1 Q2Q2 Q3Q3 - A r1r1 r3r3 r2r2 Potential is + or – based on sign of the charges Q.