Question

2. (6 points) Particles of charge +65, +48, and -95 µC and are placed in a line (Figure). The center one is 0.35 m from each of the others. Calculate the net force on each charge due to the other two. -95 μC +65 μC +48 μC 0.35 m 0.35 m 3. (3 points) Determine the magnitude and direction of the electric field at a point midway between a -8.0 µC and a +5.8 µC charge 6.0 cm apart. Assume no other charges are nearby.

Answer 1

-35 uC + 48 MC + 654C d•35 um 0:35 m ve knw by Coulonb's haw the force between is : 92 6.70 two Changes 9, aad y is the distance between the Changer E,= 8.85 x 1ó12 ka fm?/e*/sec 4,92 Where A TIE. wring tms formula we Can Calcu late force One Changeh A,B ad C. 6 Force on A: AlTE, 'AB - (65×48)X10 (65X95)X10 4īt, (0:35)? 4ūE, O70) 10१12 6175X106 13-61X1012 54. 16 x10 20 2.29 X16* - 113 x10 : 1.16 x 103 Newton The direction of the foree will be towands A . on B : For ce Atie, Y 4ūt. Yo? ( 48X95)x15 Aie Eo (a 35) 4 TiE. (0.35)2 = 2:29 x10? 3.34 x 108 - - |. 05 X105 Newton Minus bign indicaten that the for ce is direuted fromn B to C.

Force on C: fc= - - 65x95 48x 95 4īt.(D-35) 8 – 3. 34 X103 4ūG, (0-)2 = -1·13 x10 - 4.47 X10° Meuton. The Arce is directed touards C (regative bign) + 5.8luC 3. -8.0 uc 6.0 Cen The elubic fuld is depined an foreu per lenit Change at point and is 2: is i'th chanye do here the Changer are seperated by 6.0 Cam . midpoine is So the electric Hus 3.o Cen away frm each Change at midpoint is : 4τετ" - 8.0 x 10 6 4īTE. (3-0X10 2)" - 8.0 x 10 5.8 x 156 4TTE, (3-0x10-2) S.8 X156 10-13- 10-13 (- 8.0 + 5.8) X 10* 0 - 2 -2 x 107 Newton / Coulons The electric fela is direted towards Change 4,(-84nc).
Coulomb's law and the formula for electric field has been used..