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Schach [20]
2 years ago
14

Which statement best explains the relationship between the electric force between two charged objects and the distance between t

hem? A. As the distance increases by a factor, the electric force increases by the square of that factor. B. As the distance increases by a factor, the electric force increases by twice that factor. C. As the distance increases by a factor, the electric force decreases by twice that factor. D. As the distance increases by a factor, the electric force decreases by the same factor. E. As the distance increases by a factor, the electric force decreases by the square of that factor.
Physics
1 answer:
Anton [14]2 years ago
8 0

Answer: I think it’s C but I’m not completely sure about it please correct me if I’m wrong

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A thin, metallic spherical shell of radius 0.347 m0.347 m has a total charge of 7.53×10−6 C7.53×10−6 C placed on it. A point cha
USPshnik [31]

Answer:

E = 12640.78 N/C

Explanation:

In order to calculate the electric field you can use the Gaussian theorem.

Thus, you have:

\Phi_E=\frac{Q}{\epsilon_o}

ФE: electric flux trough the Gaussian surface

Q: net charge inside the Gaussian surface

εo: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2

If you take the Gaussian surface as a spherical surface, with radius r, the electric field is parallel to the surface anywhere. Then, you have:

\Phi_E=EA=E(4\pi r^2)=\frac{Q}{\epsilon_o}\\\\E=\frac{Q}{4\pi \epsilon_o r^2}

r can be taken as the distance in which you want to calculate the electric field, that is, 0.795m

Next, you replace the values of the parameters in the last expression, by taking into account that the net charge inside the Gaussian surface is:

Q=7.53*10^{-6}C+3.65*10^{-6}C=1.115*10^{-5}C

Finally, you obtain for E:

E=\frac{1.118*10^{-5}C}{4\pi (8.85*10^{-12C^2/Nm^2})(0.795m)^2}=12640.78\frac{N}{C}

hence, the electric field at 0.795m from the center of the spherical shell is 12640.78 N/C

3 0
2 years ago
A uniform electric field with a magnitude of 5750 N/C points in the positive x direction. Find the change in electric potential
castortr0y [4]

Explanation:

Given that,

Electric field = 5750 N/C

Charge q=+10.5\times10^{-6}\ C

Distance = 5.50 cm

(a). When the charge is moved in the positive x- direction

We need to calculate the change in electric potential energy

Using formula of electric potential energy

\Delta U=-W

\Delta U=-F\cdot d

\Delta U=-q(E\cdot d)

Put the value into the formula

\Delta U=-10.5\times10^{-6}\times5750\times5.50\times10^{-2}

\Delta U=-3.32\times10^{-3}\ J

The change in electric potential energy  is -3.32\times10^{-3}\ J

(b). When the charge is moved in the negative x- direction

We need to calculate the change in electric potential energy

Using formula of electric potential energy

\Delta U=-W

\Delta U=-F\cdot (-d)

\Delta U=-q(E\cdot (-d))

Put the value into the formula

\Delta U=-10.5\times10^{-6}\times5750\times(-5.50\times10^{-2})

\Delta U=3.32\times10^{-3}\ J

The change in electric potential energy  is 3.32\times10^{-3}\ J

Hence, This is the required solution.

3 0
3 years ago
Earth's gravitational force just got three times stronger! What happens to your weight?
kicyunya [14]

Your "weight" is the name you give to that gravitational force.
So your question actually says:

           "Your weight just got three times stronger !
             What happens to your weight ?"

8 0
2 years ago
State the law of conservation of machine​
laiz [17]

Answer:

law of conservation of energy is that a perpetual motion machine of the first kind cannot exist, that is to say, no system without an external energy supply can deliver an unlimited amount of energy to its surroundings

Explanation:

Hope it helps.

Mark me as Brainliest plz!

3 0
2 years ago
How much heat is absorbed by a 28g iron skillet when its temperature rises from 7oC to 29oC?
prohojiy [21]
From the information given above, 
Mass [M] = 28 g
Change in temperature = 29 - 7 = 22 
Specific heat of iron = 0.449 [This value is constant]
The formula for calculating heat absorbed, Q is
Q = Mass * Specific heat of Iron * change in temperature
Q = 28 * 0.449 * 22 = 276.58 J<span />
5 0
3 years ago
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