SECTION 27-4 1/6 13. Two equal and opposite charges of magnitude 2.0 x 10-7 C ar
ID: 1513983 • Letter: S
Question
SECTION 27-4 1/6 13. Two equal and opposite charges of magnitude 2.0 x 10-7 C are 15 arges are aranged in an equilateral triangle as in Fig. 27-16. What is Three ' C are 15 cm idway be- tion would act on apart. (a) What are the magnitude and direction of E at a point m tween the charges? (b) What force (magnitude and direction) would act an electron placed there? Answer: (a) 6.4 × 10s NC, toward the negative charge. (b) 1.0 x 10-13 N ve char tow 14. Two point charges of magnitude +2.0 x 107C and+8.5 x 10-Care 12 cm figure 27-16 apart. la) What electric field does each produce at the site of the other? b) What force acts on each Problem 12 15 arges of unknown magnitude and sign are placed a distance point apart. (a) If it is possible to have E-0 at any point not between the charges but on the line joining them, what are the necessary conditions and where is the point located (b) Is it possible, for any arrangement of two point charges, to find two points (neither at infinity) at which E-0, if so, under what conditions? Answer: (a, Charges must be opposite in sign, the nearer charge being + +- smaller in magnitude than the farther charge. (b) No. hich the electric field is figure 27-17 Problem 16 16. (a) In Fig. 27-17 locate zero. (bl Sketch qualitatively the lines of force. Take a 50 cm.Explanation / Answer
13) i) Ef = 2q/4pieor^2
Ef = (2*2*10^-7*9*10^9)/(0.15^2)
Ef = 6.4*10^5 NC^-1 (towards -q)
ii) force on an electron of charge -e
F = -Ee
F = - 6.4*10^5*2*10^-7
F = 1.02*10^-13 N (towards +q)
14) E1 = q1/4pieor2
E1 = (2.0*10^-7*9*10^9)/((0.12)^2)
E1 = 1.25*10^5 NC^-1
E2 = q2/4pieor2
E2 = (8.5*10^-8*9*10^9)/((0.12)^2)
E2 = 5.3*10^4 NC^-1
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