Answer :
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:
[tex]Q = (2.6\times 10^4) \times 1.6\times 10^{-19} = 4.16 \times 10^{-15}~C[/tex]
The electric field can be found by the following formula
[tex]E = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}[/tex]
where 'r' can be calculated as
[tex]r = \sqrt{(2\times 10^{-3})^2 + (1\times 10^{-3})^2} = 0.0022~m\\r^2 = 4.84\times 10^{-6}[/tex]
Finally, the electric field at the position (2.00 mm, 1.00 mm) is
[tex]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[/tex]
The magnitude of the electric field it produces at the position is 7.5 N/C.
The given parameters:
- Number of excess electron, n = 2.6 x 10⁴
- Position of the excess electron, x = (2.00 mm, 1.00 mm)
The position of the charged object is calculated as follows;
[tex]r^2 = (2.0 \times 10^{-3})^2 + (1.0 \times 10^{-3})^2\\\\r^2 = 5\times 10^{-6} \ m^2[/tex]
The charge of the electron is calculated as follows;
[tex]Q = nq\\\\Q = 2.6 \times 10^4 \times 1.6\times 10^{-19}\\\\Q =4.16 \times 10^{-15} \ C[/tex]
The magnitude of the electric field it produces at the position is calculated as follows;
[tex]E = \frac{F}{Q}= \frac{kQ}{r^2} = \frac{9\times 10^9 \times 4.16 \times 10^{-15}}{5\times 10^{-6}} \\\\E = 7.5 \ N/C[/tex]
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