58.5 million excess electrons are inside a closed surface. What is the net electric flux through the surface?

1 answer:

8 0

The biggest thing you're doing wrong is ignoring the units
when you're working with the quantities.

Now let's look at the rest of the problem:

The formula you used is correct:

   Net flux through the surface = (net charge inside) / ε₀

and      ε₀ = 8.85 x 10⁻¹² farad/meter.

What's the net charge inside the surface in this problem ?

It's (5.85 x 10⁷ electrons) x (the charge on each electron)

   = (5.85 x 10⁷ electrons) x (-1.6 x 10⁻¹⁹ coulomb/electron)

   =    -9.36 x 10⁻¹² coulomb .

Finally,    (net charge inside) / ε₀

   = (-9.36 x 10⁻¹² coulomb) / (8.85 x 10⁻¹² farad/meter)

   = -1.058 newton-m²/coulomb .

The sign and the significant figures in your answer are correct, so
we can see that you know what you're doing. The only thing left is
the order of magnitude. You most likely took one of the negative
exponents and made it positive. You got an answer that's 10²² too
small. Big deal. You could claim "that's close", and see whether you
can convince a teacher.

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The density of helium gas at 0.0 ◦C is 0.16 kg/m3 kg/m3. The temperature is then raised to 100.6 ◦C, but the pressure is kept co

Svetach [21]

Answer:

0.11694 kg/m³

Explanation:

T₁ = 0.0 °C = 273.15 K

ρ₁ = Density = 0.16 kg/m³

T₂ = 100.6 °C = 373.72 °C

PV = mRT

Realtion with density

$Pm=\rho RT\\\Rightarrow \rho =\frac{Pm}{RT}\\\Rightarrow \rho T= \frac{Pm}{R}$

P, m and R are constants, so

$\rho_1 T_1=\rho_2 T_2\\\Rightarrow \rho_2=\frac{\rho_1 T_1}{T_2}\\\Rightarrow \rho_2=\frac{0.16\times 273.15}{373.72}\\\Rightarrow \rho_2=0.11694\ kg/m^3$