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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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<h2>Answer:</h2>

-310J

<h2>Explanation:</h2>

The change in internal energy (ΔE) of a system is the sum of the heat (Q) and work (W) done on or by the system. i.e

ΔE = Q + W       ----------------------(i)

If heat is released by the system, Q is negative. Else it is positive.

If work is done on the system, W is positive. Else it is negative.

<em>In this case, the system is the balloon and;</em>

Q = -0.659kJ = -695J    [Q is negative because heat is removed from the system(balloon)]

W = +385J  [W is positive because work is done on the system (balloon)]

<em>Substitute these values into equation (i) as follows;</em>

ΔE = -695 + 385

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Therefore, the change in internal energy is -310J

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<em />

<em />

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