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

The distance to the center of the square is (2/2)(11.2 cm) = 0.0792 m

An electron weighs (9.11 * 10-31 kg)(9.81 m/s2) = 8.94 * 10-30 N

Each of the four charges must supply a quarter of that force, or 2.23 * 10-30 N, and that is the vertical component of a force for which the ratio of vertical to horizontal component is 11 nm / .0792 m, therefore the actual Coulomb force between one of the corners and the electron has to be 1.61 * 10-23 N.

Coulomb's law says F = kq1q2/r2; we know q2 = 1.60 * 10-19 C, we seek q1. Solving for q1,

q1 = Fr2/kq2 = (1.61 * 10-23 N)(0.0792 m)2/((8.99 * 109 Nm2/C2)(1.60 * 10-19 C))

= 7.01 * 10-17 C which is the first part of the answer; this is equal to the charge of about 437 electrons, the second part.

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