1. Suppose a speck of dust has 1.00000x1012 protons in it and a net charge of -1

Question

1. Suppose a speck of dust has 1.00000x1012 protons in it and a net charge of -17.65 nC. How many electrons does it have? nano n=10-9 1. How far apart must two charges of 96.75 nC be to have a force of 7.8 N between them?

1. A test charge of 2.84 C is placed half way between a charge of 7.4 C and 6.06 C separated by 10 cm. What is the magnitude of the total force on the test charge?

1. Find the magnitude of the electric field that exerts a force 4.66 on a proton in a hydrogen atom. Protons: q=+e e = 1.6X10-19 C

1. Three charges are located on the x-axis. Charge 4.7 C at x=3.00 cm, 4.7 C at x=11.0 cm and 2.64 C at x=8.00 cm. What is the magnitude of the force on the 2.64 C charge?

Explanation / Answer

1. Since speck of dust has -17.65nC net charge, it must have a number of electrons = number of protons + number of electrons consisting -17.65 n C charge

= 1.00000 x 10^12 + (17.65 x 10^-9)/(1.6 x 10^-19)

= 1.1103125 x 10^12 electrons

1. F= kq1 q2/r^2

r = (k q1 q2/F)^0.5

r = (8.99 x 10^9 x 96.75 x 10^-9 x 96.75 x 10^-9/ 7.8)^0.5

r = 3.28 x 10^-3 m

1. F_total = force due to 7.4 x 10^-6 C + force due to 6.06 x 10^-6 C

F_total = (8.99 x 10^9 x 2.84 x 10^-6 x 7.4 x 10^-6 / 0.05^2) - (8.99 x 10^9 x 2.84 x 10^-6 x 6.06 x 10^-6 / 0.05^2)

F_total = 13.68 N

1. F = qE

E = F/q = 4.66/ 1.6 x 10^-19 = 2.9 x 10^19 N/C

1. F_total = force due to 4.7 x 10^-6 C placed at x = 0.03 m + force due to 4.7 x 10^-6 C placed at x = 0.11 m

F_total = (8.99 x 10^9 x 2.64 x 10^-6 x 4.7 x 10^-6 / 0.05^2) - (8.99 x 10^9 x 2.64 x 10^-6 x 4.7 x 10^-6 / 0.03^2)

F_total = -79.3 N

Magnitude of F_total = 79.3 N

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