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mrs_skeptik [129]
3 years ago
12

A thin metallic spherical shell of radius 0.357 m has a total charge of 5.03 times 10^-6 C placed on it. At the center of the sh

ell is placed a point charge of 4.15 times 10^-6 C. what is the electric field at a distance of 0.815 m from the center of the spherical shell? Gauss's law states that the total electric flux contained in an enclosed area (in this case, a shell of radius 0.815m from the center of the point charge) is proportional to the total charge within the enclosed area. Thus, the electric field at the point of interest is the sum of the electric field due to the point charge and the electric field of the distributed charge treated as a point charge located at the center of the spherical shell.
Physics
1 answer:
Artyom0805 [142]3 years ago
4 0

Answer:

The electric field is 5.623\times10^{4}\ N/C

Explanation:

Given that,

Radius = 0.357 m

Charge Q=5.03\times10^{-6}\ C

Point charge q=4.15\times10^{-6}\ C

Distance = 0.815 m

We need to calculate the total electric field

Using formula of electric field

E=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{q}{r^2}

Where, q = point charge

r = distance

Put the value into the formula

E=\dfrac{9\times10^{9}\times4.15\times10^{-6}}{(0.815)^2}

E=5.623\times10^{4}\ N/C

Hence, The electric field is 5.623\times10^{4}\ N/C

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A total charge of 9.0 mC passes through a cross-sectional area of a nichrome wire in 3.6s. The number of electrons passing throu
Setler79 [48]
<h2>Given :</h2>

  • total charge = 9.0 mC = 0.009 C

Each electron has a charge of :

1.6 \times 10 {}^{ - 19} \:  C

For producing 1 Cuolomb charge we need :

  • \mathrm{\dfrac{1}{1.6 \times 10 {}^{ - 19} } }

  • \dfrac{10 {}^{19} }{1.6}

  • \dfrac{10\times 10 {}^{19} }{16}

  • \dfrac{100 \times 10 {}^{18} }{16}

  • \mathrm{6.24 \times 10 {}^{18}  \:  \: electrons}

Now, for producing 0.009 C of charge, the number of electrons required is :

  • 0.009 \times 6.24 \times  {10}^{18}

  • 0.05616 \times 10 {}^{18}

  • \mathrm{5.616 \times 10 {}^{16}  \:  \: electons}

_____________________________

So, Number of electrons passing through the cross section in 3.6 seconds is :

\mathrm{5.616 \times 10 {}^{16} \:  \: electrons}

Number of electrons passing through it in 1 Second is :

  • \dfrac{5.616 \times  {10}^{16} }{3.6}

  • \mathrm{1.56 \times 10 {}^{16}  \:  \: electrons}

Now, in 10 seconds the number of electrons passing through it is :

  • 10 \times  \mathrm{1.56 \times 10 {}^{16}  \:  \: }

  • \mathrm{1.56 \times 10 {}^{17}  \:  \: electrons}

_____________________________

\mathrm{ \#TeeNForeveR}

6 0
3 years ago
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