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djverab [1.8K]
3 years ago
9

A positive charge is placed at rest at the center of a region of space in which there is a uniform electric field. (A uniform fi

eld is one whose strength and direction are the same at all points within the region.) What happens to the electric potential energy of the positive charge, after the charge is released from rest in the uniform electric field?A. It will remain constant because the electric field is uniform.B.It will decrease because the charge will move in the opposite direction of the electric field.C.It will remain constant because the charge remains at rest.D.It will decrease because the charge will move in the direction of the electric field.E.It will increase because the charge will move in the direction of the electric field.
Physics
1 answer:
Harlamova29_29 [7]3 years ago
7 0

Answer:

.D.It will decrease because the charge will move in the direction of the electric field

Explanation:

<em>which there is a uniform electric field. (A uniform field is one whose strength and direction are the same at all points within the region.) What happens to the electric potential energy of the positive charge, after the charge is released from rest in the uniform electric field?A. It will remain constant because the electric field is uniform.B.It will decrease because the charge will move in the opposite direction of the electric field.C.It will remain constant because the charge remains at rest.D.It will decrease because the charge will move in the direction of the electric field.E.It will increase because the charge will move in the direction of the electric field.</em>

solution

The potential energy decreases, converted to kinetic energy

The charge will feel a force in the direction of the electric field (F=Eq) and thus it will accelerate with

a constant acceleration. (Just like releasing an object above the earth's surface - constant acceleration,

at least until it hits something.)

tus te answer will be

.D.It will decrease because the charge will move in the direction of the electric field

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Drupady [299]

Answer:

The magnitude of the average induced emf is 90V

Explanation:

Given;

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number of turns, N = 15 turns

magnitude of the magnetic field, B = 0.75 T

time of change of magnetic field, t = 0.05 s

The magnitude of the average induced emf is given by;

E = -NAB/t

E = -(15 x 0.4 x 0.75) / 0.05

E = -90 V

|E| = 90 V

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Suppose that the current in the solenoid is i(t). The self-inductance L is related to the self-induced EMF E(t) by the equation
Artemon [7]

Answer:

L =   μ₀ n r / 2I

Explanation:

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        E_{L} = - L dI / dt

 

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       E = - d Ф_B / dt

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        - d Ф_B / dt = - L dI / dt

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we can remove the differentials

      n Ф_B = L I

magnetic flux is defined by

     Ф_B = B . A

in this case the direction of the magnetic field is along the coil and the normal direction to the area as well, therefore the scalar product is reduced to the algebraic product

      n B A = L I

the loop area is

      A = π R²

     

we substitute

       n B π R² = L I                    (1)

To find the magnetic field in the coil let's use Ampere's law

        ∫ B. ds = μ₀ I

where B is the magnetic field and s is the current circulation, in the coil the current circulates along the length of the coil

           s = 2π R

we solve

              B 2ππ R =  μ₀ I

              B =  μ₀ I / 2πR

we substitute in

       n ( μ₀ I / 2πR) π R² = L I

       n  μ₀ R / 2 = L I

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4 0
3 years ago
The filament of a certain lamp has a resistance that increases linearly with temperature. When a constant voltage is switched on
alekssr [168]

Answer:

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Explanation:

From the question we  are told that

   The temperature coefficient is  \alpha  =  4 * 10^{-3 }\  k^{-1 }

The resistance of the filament is mathematically represented as

           R  =  R_o [1 + \alpha  \Delta T]

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Making the change in temperature the subject of the formula

     \Delta T = \frac{1}{\alpha } [\frac{R}{R_o} - 1 ]

Now from ohm law

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Substituting this we have

       \Delta T  = \frac{1}{\alpha } [\frac{I_o}{I} - 1 ]

From the question we are told that

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Substituting this we have

   \Delta T  = \frac{1}{\alpha } [\frac{I_o}{\frac{I_o}{8} } - 1 ]

=>     \Delta T  = \frac{1}{3.9 * 10^{-3}} (8 -1 )

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