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3 years ago

If the charge remains the same but the radius of the sphere is doubled, the electric flux coming out of it will be

Physics

1 answer:

il63 [147K]3 years ago

4 0

Answer:

Explanation:

We shall apply Gauss's theorem for electric flux to solve the problem . According to this theorem , total electric flux coming out of a charge q can be given by the following relation .

∫ E ds = q / ε

Here q is assumed to be enclosed in a closed surface , E is electric intensity on the surface so

∫ E ds represents total electric flux passing through the closed surface due to charge q enclosed in the surface .

This also represents total flux coming out of the charge q on all sides .

This is equal to q / ε where ε is a constant called permittivity which depends upon the medium enclosing the charge . For air , its value is 8.85 x 10⁻¹² .

If charge remains the same but radius of the sphere enclosing the charge is doubled , the flux coming out of charge will remain the same .

It is so because flux coming out of charge q is q / ε . It does not depend upon surface area enclosing the charge . It depends upon two factors

1 ) charge q and

2 ) the permittivity of medium ε around .

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What's 600,000,000 divided by 3,000.000,000,000?

Murljashka [212]

Answer:0.000002

Explanation: I Looked It Up lol

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3 years ago

In this circuit the battery provides 3 V, the resistance R1 is 7 Ω, and R2 is 5 Ω. What is the current through resistor R2? Give

sveta [45]

Answer:

The current pass the $R_2$ is $I = 0.25 A$

Explanation:

The diagram for this question is shown on the first uploaded image

From the question we are told that

The voltage is $V = 3V$

The first resistance is $R_1 = 7 \Omega$

The second resistance is $R_2 = 5 \Omega$

Since the resistors are connected in series their equivalent resistance is

$R_{eq} = R_1 +R_2$

Substituting values

$R_{eq} = 7 + 5$

$R_{eq} = 12 \Omega$

Since the resistance are connected in serie the current passing through the circuit is the same current passing through $R_2$ which is mathematically evaluated as

$I = \frac{V}{R_{eq}}$

Substituting values

$I = \frac{3}{12}$

$I = 0.25 A$

3 0

3 years ago

PART 2 OF ENERGY AND FORCES UNIT TEST

katrin2010 [14]

Answer:

1. at least two charged interacting parts

2. from the electric fields of charged subatomic particles

3 an arrow released from the bow

4Electrical fields of charged particles interact, bonding those with opposite charges.

5 the interaction of the electric fields of protons and electrons

6 The energy stored in the system increases.

7 Kinetic energy increases because the magnets move in the direction of the field.

8 Iron pieces accelerate toward the magnet, and the energy stored in the system decreases.

The energy stored in the field decreases because the magnet moves in the direction of the field.

10 The energy stored increases and then decreases.

11 The wire was not connected to the source.

12 The electromagnet will become more powerful.

the rest are written, hope this helps (:

4 0

3 years ago

(a) Two ions with masses of 4.39×10^−27 kg move out of the slit of a mass spectrometer and into a region where the magnetic fiel

sammy [17]

Answer:

Part a)

$R_1 = 0.072 m$

Part b)

$R_2 = 0.036 m$

Part c)

d = 0.072 m

Explanation:

Part a)

As we know that the radius of the charge particle in constant magnetic field is given by

$R = \frac{mv}{qB}$

now for single ionized we have

$R_1 = \frac{(4.39\times 10^{-27})(7.92 \times 10^5)}{(1.6 \times 10^{-19})(0.301)}$

$R_1 = 0.072 m$

Part b)

Similarly for doubly ionized ion we will have the same equation

$R = \frac{mv}{qB}$

$R_2 = \frac{(4.39\times 10^{-27})(7.92 \times 10^5)}{(3.2 \times 10^{-19})(0.301)}$

$R_2 = 0.036 m$

Part c)

The distance between the two particles are half of the loop will be given as

$d = 2(R_1 - R_2)$

$d = 2(0.072 - 0.036)$

$d = 0.072 m$

6 0

3 years ago

Three uniform spheres of radius 2R, R, and 3R are placed in a line, in the order given, so their centers are lined up and the sp

kolezko [41]

Answer:

x = 2.33 R from the center of mass of the smallest sphere.

Explanation:

Due to the symmetry of the spheres, the center of mass of any of them, is located just in the center of the sphere.

If we align the centers of the spheres with the x-axis, the center of mass of any of them will have only coordinates on the x-axis, so the center of mass of the system will have coordinates on the x-axis only also.

By definition, the x-coordinate of the center of mass of a set of discrete masses m₁, m₂, m₃, can be calculated as follows:

$Xcm = \frac{m1*x1+m2*x2+m3*x3}{m1+m2+3}$

In this case, we need to get the coordinates of the center of mass of each sphere:

If we place the spheres in such a way that the center of the first sphere has the x-coordinate equal to its radius (so it is just touching the origin), we will have:

x₁ = 2*R

For the second sphere, the center will be located at a distance equal to the diameter of the first sphere plus its own radius, as follows:

x₂ = 4*R + R = 5*R

Finally, for the third sphere, the center will be located at a distance equal to the diameter of the first sphere, plus the diameter of the second sphere, plus its own radius, as follows:

x₃ = 4*R + 2*R + 3*R = 9*R

We can calculate the mass of each sphere (assuming that all are from the same material, with a constant density), as the product of the density and the volume:

m = ρ*V

For a sphere, the volume can be calculated as follows:

$\frac{4}{3} *\pi *(r)^{3}$

So, we can calculate the masses of the spheres, as follows:

m₁ = ρ* $\frac{4}{3} *\pi *(2r)^{3}$

m₂ = ρ* $\frac{4}{3} *\pi *(r)^{3}$

m₃ = ρ* $\frac{4}{3} *\pi *(3r)^{3}$

The total mass can be calculated as follows:

M= ρ* $\frac{4}{3} *\pi$ * (8*r³ + r³ + 27*r³) =ρ* $\frac{4}{3} *\pi$ * 36*r³

Replacing by the values, and simplifying common terms, we can calculate the x-coordinate of the center of mass of the system as follows:

$Xcm = \frac{m1*x1+m2*x2+m3*x3}{m1+m2+3}$

$Xcm = \frac{(8*R^{3} *2*R)+(R^{3}*(5*R))+27*R^{3}*(9*R))}{36*R^{3} }=\frac{264*R^{4} x}{36*R^{3}} = 7.33 R$

As the x-coordinate of the center fof mass of the entire system is located at 7.33*R from the origin, and the center of mass of the smallest sphere is located at 5*R from the origin, the center of mass of the system is located at a distance d:

d = 7.33*R - 5*R = 2.33 R

4 0

3 years ago