## 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

Question

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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1 year 2021-09-04T12:57:21+00:00 2 Answers 8 views 0

1. Explanation:

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$$3.9*10^{5}N/C$$

Explanation:

from the expression for determing the magnetic field strength

$$E=\frac{q}{4\pi e_{0}d^2 }\\$$

since the charge is given as

$$q=1.56*10^{-5}c\\$$

and the distance is

d=60cm=0.6m

We can calculate the constant k

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

if we substitute values, we arrive at

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