In Figure 17.40(b), q = 1.00 µC, b = 5.00 and c = 6.00. Figure 17.40(b) Enter \'

Question

In Figure 17.40(b), q = 1.00 µC, b = 5.00 and c = 6.00. Figure 17.40(b) Enter 'infinity' for the following answers where appropriate.

(a) Find the total electric field at x = 5.00 cm. N/C

(b) Find the total electric field at x = 12.0 cm. N/C

(c) If the charges are allowed to move and eventually be brought to rest by friction, what will the final charge configuration be? (That is, will there be a single charge, double charge, etc., and what will its value(s) be? List the charges from left to right if there are more than one, using the format '+3q,+q'. Include the sign for each charge.

Explanation / Answer

A) Infinity.

B) Net electric field, Enet = E1 + E2 + E3 + E4

E = kq/r^2

E1 = 9*10^9*5*10^-6 / 0.12^2 = 3.125*10^6 (-i) N/C

E2 = 9*10^9*10^-6 / 0.08^2 = 1.40625*10^6 (i) N/C

E3 = 9*10^9*6*10^-6 / 0.05^2 = 2.16*10^7 (i) N/C

E4 = 9*10^9*10^-6 / 0.01^2 = 9*10^7 (i) N/C

So, Enet = -3.125*10^6 + 1.40625*10^6 + 2.16*10^7 + 9*10^7 i N/C

Enet = 1.0988 * 10^8 N/C

C) final charge configuration will be equal to 3q - 2(q) = +q

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

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