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Ostrovityanka [42]
3 years ago
10

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, t

he 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.

Physics
2 answers:
Serga [27]3 years ago
5 0

Explanation:

Below is an attachment containing the solution.

Nastasia [14]3 years ago
3 0

Answer

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

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Answer:

The answer is \frac{F_{Sun-Mercury} }{F_{Sun-Earth} } =0,3709. Let's learn why.

Explanation:

Newton's law of universal gravitation says;

F_{g} =G.\frac{m_{1}.m_{2}}{r^{2}}

Here G is a universal gravitational <u>constant</u> and is measured experimentally.

Sun's gravitational pull on mercury is:

F_{Sun-Mercury} =G.\frac{m_{sun}.3,30.10^{23}}{(5,79.10^{10})^{2} }

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As a result;

\frac{F_{Sun-Mercury}}{F_{Sun-Earth} }=\frac{Gm_{sun}98,4366}{Gm_{sun}265,33 } =0,3709

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Answer:

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