Figure a below shows two charged particles fixed in place on an x axis with sepa

Question

Figure a below shows two charged particles fixed in place on an x axis with separation L. The ratio q1/q2 of their charge magnitudes is 5.00. Figure b shows the x component Enet,x of their net electric field along the x axis just to the right of particle 2. The x axis scale is set by xs = 36.0 cm. (a) At what value of x > 0 is Enet,x maximum? (b) If particle 2 has charge ?q2 = ?3e, what is the value of that maximum?

Explanation / Answer

Part A Let the magnitude of negative charge be q. That of positive charge separated L cm away will be 9q The net electric field at any point x > 0 is given by Enet = (9*10^9)q*(10^4)*[9/(x+L)^2 - 1/x^2] ----------------------------- 1 To find where it will be maximum we differentiate [9/(x+L)^2 - 1/x^2] with respect to x and equate it to zero -18/(x+L)^3 + 2/x^3 = 0 or (x+L)^3 / x^3 = 9 = (x+L)/x = 9^(1/3) = 2.08 also d^2/dx^2[[9/(x+L)^2 - 1/x^2] = 54/(x+L)^4 - 6/x^4 = C say so C*(x+L)^4= 54 - 6*[(x+L)/x]^4 = 54 - 6* (2.08^4) = 54 - 112.3 for x = 2.08. So C < 0 for x = 2.08 So at x = 2.08 E net is maximum. It is given that at x = 20 cm 1 cm less than xs in the figure, the field is zero. That is possible only when 9/(20+L)^2 = 1/20^2 or (20+L)/20 = 3 or 1 + L/20 = 3 or L = 40 cm Substituting the appropriate values for q1 and q2 we get Enet(max) = (9*10^9)(3*10^-19)*(10^4)*[9/(2.08+40)^2 - 1/(2.08)^2] = (27*10^-6)*(-0.226) = -6.10*10^-6 N/C

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.