Question

Use Coulomb's law to determine the magnitude of the electric field at points A and B in (Figure 1) due to the two positive charges (Q = 4.3 μC ) shown. Suppose that a = 4.9 cm . Determine the magnitude of the electric field at point A and point B.

y B A x |-- a + +Q a a 2a

Answer 1

126.8 Ej TL < 26.60 +Q 090 26-6-> +Q kaa ► 2a a= 4.9 cm = 0.049m Q = 4.340 6.3x1öbe r == (2a) 4 (a)? = 1(2469)2+(4.972 =10.96 cm -0. Ilm Electric field at point A dueto left ta change cm Ei=k of = 9x109x 4.38106 (o. 1172 = 3.2 x 106N/ Elecmic field at point A due to right to change F2 =k = 126.8 4.3x1ās = 9x107x 10.11) = 3-2 X10N Angle between E, and Ez is Magnitude of electric field at point A is EA VE, ?+ Ez + 2 E, Ez eos - (3-2)2 + (3.2)? + 2 x3.2 x3.2 Cos 126-8**10° En = 2.87 xlo 6 NIC 2 CScanned with CamScanner

116.60 a Ž 45° 18.40 L ta tastattata 구 due E a 4.9 emo q 4.3ue =4.3xlore ătar v pao y = Vărqa rio a Electric field at B to left to change Q R: ry 9x109 x 4.3x7of 29 9 x 4.3 103 2x (0.049) NA 8.06*108 m/ Electric field at B due to right to charge Ek =9xto x 4.3x10-6 Tox (0.049)_172 = 7.61*10* E 11 B %

B Angle between 3 and E 116.6° E . Met field at B Eg +25, 5, Cos 116.69 (8.06x10)*+67x106)=0.896714 To $64.96 +2.59 - 11.63 H/ 7:48x 109 Ne