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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.

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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A capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has radius 11.0 cm ,

viktelen [127]

Part A)
First of all, let's convert the radii of the inner and the outer sphere:
$r_A = 11.0 cm = 0.110 m$
$r_B = 16.5 cm=0.165 m$
The capacitance of a spherical capacitor which consist of two shells with radius rA and rB is
$C=4 \pi \epsilon _0 \frac{r_A r_B}{r_B- r_A}=4\pi(8.85 \cdot 10^{-12}C^2m^{-2}N^{-1}) \frac{(0.110m)(0.165m)}{0.165m-0.110m}=$
$=3.67\cdot 10^{-11}F$

Then, from the usual relationship between capacitance and voltage, we can find the charge Q on each sphere of the capacitor:
$Q=CV=(3.67\cdot 10^{-11}F)(100 V)=3.67\cdot 10^{-9}C$

Now, we can find the electric field at any point r located between the two spheres, by using Gauss theorem:
$E\cdot (4 \pi r^2) = \frac{Q}{\epsilon _0}$
from which
$E(r) = \frac{Q}{4 \pi \epsilon_0 r^2}$
In part A of the problem, we want to find the electric field at r=11.1 cm=0.111 m. Substituting this number into the previous formula, we get
$E(0.111m)=2680 N/C$

And so, the energy density at r=0.111 m is
$U= \frac{1}{2} \epsilon _0 E^2 = \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(2680 N/C)^2=3.17 \cdot 10^{-5}J/m^3$

Part B) The solution of this part is the same as part A), since we already know the charge of the capacitor: $Q=3.67 \cdot 10^{-9}C$ . We just need to calculate the electric field E at a different value of r: r=16.4 cm=0.164 m, so
$E(0.164 m)= \frac{Q}{4 \pi \epsilon_0 r^2}=1228 N/C$

And therefore, the energy density at this distance from the center is
$U= \frac{1}{2}\epsilon_0 E^2 = \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(1228 N/C)^2=6.68 \cdot 10^{-6}J/m^3$

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3 years ago

A piston-cylinder device initially contains 0.08 m3 of nitrogen gas at 150 kPa and 200°C. The nitrogen is now expanded to a pres

Lemur [1.5K]

Answer:

$V_2 = 0.125 m^3$

Work done = = 5 kJ

Explanation:

Given data:

volume of nitrogen $v_1 = 0.08 m^3$

$P_1 = 150 kPa$

$T_1 = 200 degree celcius = 473 Kelvin$

$P_2 = 80 kPa$

Polytropic exponent n = 1.4

$\frac{T_2}{T_1} = [\frac{P_2}{P_1}]^{\frac{n-1}{n}$

putting all value

$\frac{T_2}{473} = [\frac{80}{150}]^{\frac{1.4-1}{1.4}$

$\frac{T_2} = 395.23 K = 122.08 DEGREE \ CELCIUS$

polytropic process is given as

$P_1 V_1^n = P_2 V_2^n$

$150\times 0.08^{1.4} = 80 \times V_2^{1.4}$

$V_2 = 0.125 m^3$

work done $= \frac{P_1 V_1 -P_2 V_2}{n-1}$

$= \frac{150 \times 0.8 - 80 \times 0.125}{1.4-1}$

= 5 kJ

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