Question

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

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

$\phi=EAcos\theta\\=3.4 *10^{23}* *2.3 *10^{-2}cos 84°\\=0.82*10^{21}Wb$

5.$\phi=EAcos\theta\\=EAcos 90°\\=0Wb$