1 answer

I have been struggling with these two practice problems in the book, thanks for the help...

Question:

1. (22.4) Richard Feynman (1918-1988) once said if two people stood at arms length from each other and each person had 1% more electrons than protons, the force of repulsion between them would be enough to lift a weight equal to that of the entire Earth. Do an order-of- magnitude calculation to show this. Be sure to communicate assumptions and estimates. 2. (23.20) A particle with charge Q is located at the center of a cube with edges of length L Also, six other identical charged particles of charge q are positioned symmetrically around Q as shown in the figure below. Assume q is a negative number. Determine the electric flux through one face of the cube.

I have been struggling with these two practice problems in the book, thanks for the help in advance!

1. (22.4) Richard Feynman (1918-1988) once said if two people stood at arm's length from each other and each person had 1% more electrons than protons, the force of repulsion between them would be enough to lift a "weight equal to that of the entire Earth. Do an order-of- magnitude calculation to show this. Be sure to communicate assumptions and estimates. 2. (23.20) A particle with charge Q is located at the center of a cube with edges of length L Also, six other identical charged particles of charge q are positioned symmetrically around Q as shown in the figure below. Assume q is a negative number. Determine the electric flux through one face of the cube.

Answers

Richard Feynman assEdition is true.

Solution

Suppose their masses are of the order 102 kg If protons constitute half of total mass - other half being neutrons, there will be the same number of electrons as protons in neutral matter. So, 1% electrons will be same as 1% protons. Protons-102 kg/ [1 .673x10-27 kg-6x1028 protons If there is a net charge of 1% of total electrons than protons, q-01-a2-(1 /100)x 6x1 028 proton charge proton charge eletron charge e 1.602x10 28 So, q = (1/10Ox 6x10 x 1.602x10-19 C-~ 10% So, using F Kq/R, where R - distance 1m, K 9x109~ 1010, q ~10° C, F~ 1026N For lifting Earth, gravitational force-Fg = Mearth x g Mearthä 6 x 1024 kg g~10 m/s2 So, Fg ~1026 N So, clearly the assertion is true!

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