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What is the magnitude of the electric field at a distance 60 cm from the center of the sphere? The radius of the sphere 30 cm, the charge on the sphere is 1.56 × 10−5 C and the permittivity of a vacuum is 8.8542 × 10−12 C 2 /N · m2 . Answer in units of N/C.

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Answer

[tex]3.9*10^{5}N/C[/tex]

Explanation:

from the expression for determing the magnetic field strength

[tex]E=\frac{q}{4\pi e_{0}d^2 }\\[/tex]

since the charge is given as

[tex]q=1.56*10^{-5}c\\[/tex]

and the distance is

d=60cm=0.6m

We can calculate the constant k

[tex]K=\frac{1}{4\pi e_{0}}\\ K=\frac{1}{4\pi *8.8542*10^{-12}}\\k=8.98*10^{9}\\[/tex]

if we substitute values, we arrive at

[tex]E=\frac{8.98*10^9 *1.56*10^{-5}}{0.6^2} \\E=389133.33\\E=3.9*10^{5}N/C[/tex]

Explanation:

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