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3 years ago

A source charge generates an electric field of 1236 N/C at a distance of 4 m. What is the magnitude of the source charge?

Physics

2 answers:

SashulF [63]3 years ago

8 0

Answer:

Explanation:

A region around a charged particle where another charged particle experiences a force is called electric field.

Electric field strength is defined as the force experienced by unit positive test cahrge when placed in an electric field.

The formula of electric field due to a point cahrge is given by

E = k q / r^2

Where k is a constant and it's value is

9 × 10^9 Nm^2/C^2.

1236 = 9 × 10^9 × q / 16

q = 1.48 × 10^-3 Coulomb

lara [203]3 years ago

6 0

The magnitude of the source is 2.2 C

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Force on a moving charge is given by

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7 0

3 years ago

Read 2 more answers

2. the dipole moment of a dipole in a 300-n/c electric field is initially perpendicular to the field, but it rotates so it is in

aleksandr82 [10.1K]

Work Done (W) by the field is-6x 107 J,

<h3>What is Electric dipole?</h3>

A pair of opposite, non-coplanar, equally powerful electric charges that are in opposition to one another. An atom is said to have a "induced electric dipole" if the center of the negative cloud of electrons has moved a little bit away from the nucleus due to an external electric field. When the external field is taken away, dipolarity is lost.

Electric field (E) = 300 N/C

Dipole moment (p) = 2 x 10° Cm

Solution:

From the formula we know.

U = -pE cosФ

Here,

p Denotes Dipole moment.

E Denotes Electric field.

Ф Denotes angle b/w them

Now, as given, firstly the dipole is perpendicular to the electric field, so

angle (Ф1) will be 90° and now the dipole is rotated such that they are in same

direction so the angle (Ф2) will be 0°

So, let's find Change in Potential energy which will be equal to the work done

by the electric field.

ΔU = Uf - Ui

ΔU = [-pE*cos Ф2] - [-pE *cos Ф1]

ΔU = [-pE*cos Ф2] + pE *cos Ф1

ΔU = pE * [cos Ф2+ cos Ф1]

Substituting the values,

ΔU = pE * [cos 0° + cos 90°]

ΔU = pE * (-1 +0)

ΔU = -pE

ΔU = -2x 10^-9 × 300

ΔU = 6 x 10^(-9+2)

ΔU = 6 x 10^-7

W = ΔU = -6 x 10^-7

W = - 6 x 10 7 J

Work Done (W) by the field is - 6 x 10-7 J.

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2 years ago

1. Determine the energy released per kilogram of fuel used.

Serga [27]

The formula for energy release per kilogram of fuel burned is energy release per kg=6.702*10-13. and 19. J 1 Mev = 1.602 X 10 T

Calculate the energy in joules per kilogram of reactants given MeV per reaction. Energy is the ability or capacity to perform tasks, such as the ability to move an item (of a certain mass) by exerting force. Energy can exist in many different forms, including electrical, mechanical, chemical, thermal, or nuclear, and it can change its form.
Think of a mole of plutonium-239 (molar mass: 239 grams) as a mole of "reactions."

Energy used in the US per person annually = 3-5 X 1011
Population (number of people) = 3.108The required mass of the fuel is 3.5x1011 x3-1x10 8x 10)/6.703 X1013 kg. the mass required: 1.62 x 1033 kg Mev in Joules 6 is equal to 101.60*I0-
19. J 1 Mev = 1.602 X 10 T, which translates to 1.602*1013/2.39x10-3 energy release per kilogram, or 6.702*10-13.

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2 years ago

James makes a treasure map to locate a secret item on McEachern Campus. He uses the

Inessa [10]

Answer:

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Explanation:

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3 years ago

Read 2 more answers

An electron with a speed of 0.95c is emitted by a supernova, where cc is the speed of light. What is the magnitude of the moment

krok68 [10]

Answer:

2.59×10¯²² Kgm/s

Explanation:

Data obtained from the question include:

Velocity of electron = 0.95c

Momentum =?

Next, we shall determine the velocity of the electron. This can be obtained as follow:

Velocity of electron = 0.95c

Velocity of Light (c) = 3×10⁸ m/s

Velocity of electron = 0.95c

Velocity of electron = 0.95 × 3×10⁸

Velocity of electron = 2.85×10⁸ m/s

Finally, we shall determine the mometum of the electron.

Momentum is simply defined as the product of mass and velocity. Mathematically, it is expressed as:

Momentum = mass x Velocity

Thus, with the above formula, we calculate the momentum of the electron as follow:

Mass of electron = 9.1×10¯³¹ Kg

Velocity of electron = 2.85×10⁸ m/s

Momentum of electron =?

Momentum = mass x Velocity

Momentum = 9.1×10¯³¹ × 2.85×10⁸

Momentum = 2.59×10¯²² Kgm/s

Therefore, the momentum of the electron is 2.59×10¯²² Kgm/s

3 0

3 years ago