Answer to Question #153236 in Electricity and Magnetism for zaina

Question

A -2.00 microC point charge of mass 8.00 g is suspended on a string and placed in a horizontal electric field. The mass is in equilibrium when the string forms a 45 angle with the vertical.

1. Find the electric force on the charge in this field

2.Find the strength of the electric field.

3.How many electrons are there in 10.00 C? How many electrons are there in

15.00 microC?

Answer 1

Solution

$1.\;\frac{mg}{F}=\tan{45}=1;\\ F=mg=9,8\times0.008=0.0784N;\\ 2.\;E=\frac{F}{q}=\frac{0.0784}{2\times10^{-6}}=3,92\times10^4\;\frac{V}{ m}\\3.\;N_1=\frac{Q_1}{e}=\frac{10}{1,6\times{10^{-19}}}=6,25\times{10^{19}}\\N_2=\frac{Q_2}{e}=\frac{15\times{10^{-6}}}{1,6\times{10^{-19}}}=9,375\times{10^{13}}$