Question

A small, charged, spherical object at the origin of a Cartesian coordinate system contains 2.60 × 10 4 more electrons than protons. What is the magnitude of the electric field it produces at the position (2.00 mm, 1.00 mm)?

Answer 1

Answer:

E = 7.77 N/C

Explanation:

The charge of a single electron is 1.6 x 10^{-19} C. The net charge of the object is therefore the multiplication of the number of excess electrons and the charge of a single electron:

$Q = (2.6\times 10^4) \times 1.6\times 10^{-19} = 4.16 \times 10^{-15}~C$

The electric field can be found by the following formula

$E = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}$

where 'r' can be calculated as

$r = \sqrt{(2\times 10^{-3})^2 + (1\times 10^{-3})^2} = 0.0022~m\\r^2 = 4.84\times 10^{-6}$

Finally, the electric field at the position (2.00 mm, 1.00 mm) is

$E = \frac{1}{4\pi(8.8\times 10^{-12})}\frac{4.16\times 10^{-15}}{4.84\times 10^{-6}} = 7.77~N/C$