Four point charges, q , are fixed to the four corners of a square that is 11.2 c

Question

Four point charges, q, are fixed to the four corners of a square that is 11.2 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 11 nm above the center of the square. (The square is horizontally flat, and the electron is suspended 11 nm vertically above the center of the square.) What is the magnitude of each fixed charge in coulombs?
C

What is the magnitude of each fixed charge as a multiple of the electron's charge?
e

Explanation / Answer

Please change your figures or let me know if you need further help

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.

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b) q = ne

   n = q/e = 96.66*10^-13 electrons.

therefore q = 96.66*10^-13 e

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