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Dmitry [639]
2 years ago
7

Calculate the electric field on the surface of a charged sphere with a radius of 5cm and charge 10μC. Thanks

Physics
1 answer:
victus00 [196]2 years ago
4 0

As you approach the surface of the sphere very closely, the electric field should resemble more and more the electric field from an infinite plane of charge.

If you check Gauss's law (recalling that the field in the conductor is zero) you will see that if the surface charge density is σ=Q/4πR2, then indeed the field at the surface is σ/ϵ0 as in the infinite charge of plane case.

Such a field is constant, the field lines are parallel and non-diverging, and the infinities associated with the field due to point charge do not arise.

Explanation:

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What is the strength of the electric field inside the membrane just before the action potential?
gtnhenbr [62]

Answer:

Incomplete question, check attachment for the graph needed to solve problem.

A 8.1nm........

Explanation:

Electric Field is given as

E=V/d

Where V is voltage

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The voltage V just before action of potential is -70mV,

The value d=8.1nm

d=8.1×10^-9m

E=V/d

E=-70×10^-3/8.1×10^-9

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5 0
4 years ago
a glass bottle of soda is sealed with a screw cap the absolute pressure of the gas inside the bottle is 1.80*10^5 pa. assuming t
eduard

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We can multiply both sides of the formula by S to obtain the force.

P*S=(F*S)/S

P*S=F

Solve for P=1.80*10^5 Pa and S=4.10*10^-4 m^2 ([Pa] =[N/m^s])

(1.80*10^5 N/m^s) * (4.10*10^-4 m^2) =F

73.8 N =F

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