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Sladkaya [172]
3 years ago
12

A long, solid rod 5.3 cm in radius carries a uniform volume charge density. Part A If the electric field strength at the surface

of the rod (not near either end) is 22 kN/C , what is the volume charge density?
Physics
1 answer:
Blizzard [7]3 years ago
6 0

Answer:

\rho = 7.35\times \muC/m^3

Explanation:

given,

Radius of the solid rod, R = 5.3 cm

Electric field strength,E = 22 kN/C

Let the volume charge density be ρ

From Gauss law

  E = \dfrac{Rl}{2\epsilon_0}

 ε₀ is the permitivity of free space

  R is the radius of the rod

 and also,

\rho=\dfrac{2E\epsilon_0}{R}

ρ is the volume charge density

\rho=\dfrac{2\times 22\times 10^{3}\times 8.854\times 10^{-12}}{0.053}

\rho = 7.35\times 10^{-6}\ C/m^3

\rho = 7.35\times \muC/m^3

Hence, the volume charge density is equal to \rho = 7.35\times \muC/m^3

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Hello!

<span>Let us apply the time function of space, in the uniform uniform motion (UUM)
</span>
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\boxed{\boxed{t \approx 5.3\:s}}\end{array}}\qquad\quad\checkmark

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<span>C. 5.3s</span>
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