1. Compute the force of attraction between
a +1.60 x 10 ^-19 C charge and a -2.09 x 10^-18 C charge if they are 4.01 x 10^-10 m apart.
2. Two ball bearings with opposite charges are 1.11 m apart on the floor on the floor. What are their charges if they are attracted to each other with a force of 5.11 N?
3. Compute the electric field experienced by a test charge q =+0.80μC from a source = +0.15μC in a vacuum when the test charge is places 0.20 m away from the other charge.
4. A.
E = 3.4 x 10^23 N/C
A = 2.3 x 10-2m²
Cos = 84⁰
B.
E= 8.2 x 10^5 N/C
A = 6.1 x 10^-7m²
Cos = 90⁰
Answer
1.force of attraction
"F=\\frac{kqq'}{r^2}\\\\=\\frac{9*10^9*1.6*2.09*10^{-12}}{(4.01*10^{-10})^2}\\\\=1.87*10^{17}N"
2.charge of balls is
"Q=\\sqrt{\\frac{Fr^2}{k}}\\\\=\\sqrt{\\frac{5.11*1.11*1.11}{9*10^9}}\\\\=2.65*10^{-4}C"
3.elecyric field is given on test charge is
"E=\\frac{kq}{r^2}\\\\=\\frac{9*10^9*0.15*10^{-6}}{0.2*0.2}\\\\=33.75*10^3N"
4.flux can be written
"\\phi=EAcos\\theta\\\\=3.4 *10^{23}*\n\n*2.3 *10^{-2}cos 84\u00b0\\\\=0.82*10^{21}Wb"
5."\\phi=EAcos\\theta\\\\=EAcos 90\u00b0\\\\=0Wb"
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