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Tasya [4]
3 years ago
8

Consider a spherical capacitor with radius of the inner conducting sphere a and the outer shell b. The outer shell is grounded (

i.e., it is at zero potential). The charges are +Q and −Q. a A B +Q −Q b What is the magnitude of electric field at in the region between the sphere and the outer

Physics
1 answer:
AleksAgata [21]3 years ago
6 0

Answer:

Explanation:

The application of Gauss's law is used in the derivation as shown with detailed step by step in the attached file.

The potential difference on this spherical capacitor is ΔV = Va - Vb = kQ/a - kQ/b = kQ(1/a - 1/b)

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8 0
3 years ago
Read 2 more answers
Help ASAP
EastWind [94]

Answer:

Elastic potential energy, E=2.35\times 10^{-8}\ J

Explanation:

Charge, q=9.4\times 10^{-10}\ C

Potential, V = 50 V

It is required to find the electric potential energy in a capacitor stored in it. The formula of the electric potential energy in a capacitor is given by :

E=\dfrac{1}{2}qV\\\\E=\dfrac{1}{2}\times 9.4\times 10^{-10}\times 50\\\\E=2.35\times 10^{-8}\ J

So, the electric potential energy stored in the capacitor is 2.35\times 10^{-8}\ J

8 0
3 years ago
An object has an acceleration of 6.0 m/s/s. If the net force was doubled and the mass was one-third the original value, then the
alexandr402 [8]

Hahahahaha. Okay.

So basically , force is equal to mass into acceleration.

F=ma

so when F=ma , we get acceleration=6m/s/s

Force is doubled.

Mass is 1/3 times original.

2F=1/3ma

Now , we rearrange , and we get 6F=ma

So , now for 6 times the original force , we get 6 times the initial acceleration.

So new acceleration = 6*6= 36m/s/s

5 0
3 years ago
It takes 130 J of work to compress a certain spring 0.10m. (a) What is the force constant of this spring? (b) To compress the sp
Ede4ka [16]

Explanation:

Given that,

Work done to stretch the spring, W = 130 J

Distance, x = 0.1 m

(a) We know that work done in stretching the spring is as follows :

W=\dfrac{1}{2}kx^2\\\\k=\dfrac{2W}{x^2}\\\\k=\dfrac{2\times 130}{(0.1)^2}\\\\k=26000\ N/m

(b) If additional distance is 0.1 m i.e. x = 0.1 + 0.1 = 0.2 m

So,

W=\dfrac{1}{2}kx^2\\\\W=\dfrac{1}{2}\times 26000\times 0.2^2\\\\W=520\ J

So, the new work is more than 130 J.

6 0
3 years ago
How does the use of simple machines affect force and distance when work is done?
bazaltina [42]

Answer:

Explanation:

Machines simply make work easier to do. They increase the amount of force exerted on a body and also the distance through which the force is applied. Also, they can also change the direction through which force on them is applied in order to produce much more work.

        Work done = force x distance

The input force in a machine is attenuated to yield even more force. This is the purpose of designing a simple machine. When the force increases, more work would be produce with our little effort applied on the body.

Work done is a function of the force applied on a body and the distance through which it moves.

8 0
3 years ago
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