Four point charges, q, are fixed to the four corners of a square that is 14.9 cm
ID: 2163364 • Letter: F
Question
Four point charges, q, are fixed to the four corners of a square that is 14.9 cm on a side. An electron is suspended above a point at which its weight is balanced by the electrostatic force due to the four point charges, at a distance of 19 nm above the center of the square. (The square is horizontally flat, and the electron is suspended 19 nm vertically above the center of the square.) What is the magnitude of each fixed charge in coulombs?What is the magnitude of each fixed charge as a multiple of the electron's charge?
Explanation / Answer
first we find the angle , Tan? =12*10^-9/0.08343 ? = 8.240*10^-6 degree the resultant force F' = 4Fsin? since the horizontal components cancel each other, we added vertical components. here F = kq(e)/r^2 here e = 1.602*10^-19 C ( charge of the electron ) therefore the resultant force F' = 4*kq(e)/r^2*sin? but F = mg mg = 4*kq(e)/r^2*sin? q = mg*r^2 sin?/4*k*(e) = 9.1*10^-31*9.8*r^2*1.4381*10^-7/(4*9*10^9*1.602*10^-19) = 154.86*10^-32 C ( since sin? = 1.4381*10^-7 and r^2 = 69.64*10^-4 m ) the magnitude of each fixed charge is 154.86*10^-32 C. ------------------------------------ b) q = ne n = q/e = 96.66*10^-13 electrons. therefore q = 96.66*10^-13e
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.