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Vlad [161]
3 years ago
5

The electric field everywhere on the surface of a thin, spherical shell of radius 0.770 m is of magnitude 860 N/C and points rad

ially toward the center of the sphere.
(a) What is the net charge within the sphere's surface?
(b) What can you conclude about the nature and distribution of the charge inside the spherical shell?
Physics
1 answer:
11111nata11111 [884]3 years ago
5 0

Answer:

(a) Q = 7.28\times 10^{14}

(b) The charge inside the shell is placed at the center of the sphere and negatively charged.

Explanation:

Gauss’ Law can be used to determine the system.

\int{\vec{E}} \, d\vec{a} = \frac{Q_{enc}}{\epsilon_0}\\E4\pi r^2 = \frac{Q_{enc}}{\epsilon_0}\\(860)4\pi(0.77)^2 = \frac{Q_{enc}}{8.8\times 10^{-12}}\\Q_enc = 7.28\times 10^{14}

This is the net charge inside the sphere which causes the Electric field at the surface of the shell. Since the E-field is constant over the shell, then this charge is at the center and negatively charged because the E-field is radially inward.

The negative charge at the center attracts the same amount of positive charge at the surface of the shell.

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Explanation:

assuming ideal gas behaviour of the gas , the equation for ideal gas is

P*V=n*R*T

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the work that is done by the gas is calculated through

W=∫pdV=  ∫ (n*R*T/V) dV

for an isothermal process T=constant and since the piston is closed vessel also n=constant during the process then denoting 1 and 2 for initial and final state respectively:

W=∫pdV=  ∫ (n*R*T/V) dV =  n*R*T  ∫(1/V) dV = n*R*T * ln (V₂/V₁)

since

P₁=n*R*T/V₁

P₂=n*R*T/V₂

dividing both equations

V₂/V₁ = P₁/P₂

W= n*R*T * ln (V₂/V₁)  = n*R*T * ln (P₁/P₂ )

replacing values

P₁=n*R*T/V₁ = 2 moles* 8.314 J/mol K* 300K / 0.1 m3= 49884 Pa

since P₂ = 1 atm = 101325 Pa

W= n*R*T * ln (P₁/P₂ ) = 2 mol * 8.314 J/mol K * 300K * (49884 Pa/101325 Pa) = -3534.94 J

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The vector perpendicular to the plane of A = 3i+ 6j - 2k and B = 4i-j +3k is 16 i - 17 j - 27 k

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brainly.com/question/14384780

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