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muminat
3 years ago
6

A positive charge moves in the direction of an electric field. Which of the following statements are true? Check all that apply.

Check all that apply. The potential energy associated with the charge increases. The amount of work done on the charge cannot be determined without additional information. The electric field does positive work on the charge. The electric field does not do any work on the charge. The potential energy associated with the charge decreases. The electric field does negative work on the charge. SubmitRequest Answer Provide Feedback Next
Physics
2 answers:
Sergio039 [100]3 years ago
5 0

Answer:

The electric field does positive work on the charge.

The potential energy associated with the charge decreases.

Explanation:

We know that force due to electric filed on a charge particle q given as

 F = q E          --------1

If the   positive charge is moving in the direction of electric filed it means that force due to electric field and displacement of particle due to motion is in the same direction.So the work done due to electric filed will be positive.

Work = Force . Displacement

W= F.d

F= W/d            ---------2

From equation 1 and 2

W/d = q E

W/ q = E/ d

We know that voltage difference ΔV

ΔV = E/ d

ΔV =W/ q

So the potential difference will be decrease because charge is positive.So the potential energy will be decreases.

jarptica [38.1K]3 years ago
3 0

Answer:

The potential energy associated with the charge decreases,

The electric field does positive work on the charge.

Explanation:

If a positive charge is moving in the direction of an electric field that it would normally move, the potential energy associated with the charge will decrease because the scenario is similar to what occurs in a constant gravitational field.

Also, when potential energy is lost (decreases), this means positive work is done on the charge.

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Answer:

a)   I2 = 3 (o-10) / (o- 30) , b)   h ’/h=  3 (o-10) / o (o-30)

Explanation:

The builder's equation is

          1 / f = 1 / o + 1 / i

Where f is the focal length, or e i are the distance to the object and image, respectively

As the separation between the lenses is greater than the focal distances, we must work them individually and separately. Let's start with the leftmost lens with focal length f = 15 cm

Let's calculate the position of the image of this lens

         1 / i1 = 1 / f - 1 / o

         1 / i1 = 1/15 - 1 / o

         i1 = o 15 / (o-15)

Let's calculate the distance to the image of the second lens, for this the image of the first is the distance to the object of the second

        o2 = d-i1

We write the builder equation

       1 / f2 = 1 / o2 + 1 / i2

       1 / i2 = 1 / f2 -1 / o2

       1 / i2 = 1 / f2 - 1 / (d-i1)

       1 / i2 = 1/20 - 1 / (d-i1)            (1)

Let's evaluate the last term

      d-i1 = d - 15 o / (o-15)

      d-i1 = (d (o-15) - 15 o) / (o-15)

      d- i1 = (30 or -30 15 -15 o) / (o-15)

      d-i1 = (15 or - 450) / (o- 15)

      d-i1 = = (15 or -450) / (o-15)

replace in 1

      1 / i2 = 1/20 - (or - 15) / (15 or -450)

      1 / i2 = [(15 o-450) - (o-15) 20] / (15 or -150)

      1 / i2 = (15 or - 450 - 20 or + 300) / (15 or - 150)

      1 / i2 = (-5 or -150) / (15 or -150)

      1 / i2 = (or -30) / (3 or - 30)

      I2 = 3 (o-10) / (o- 30)

Part B

The height of the image, we use the magnification equation

     m = h ’/ h = - i / o

     h ’= - h i / o

In our case

     h ’= h i2 / o

     h ’= h 3 (o-10) / o (o-30)

If they give the distance to the object it is easier

5 0
3 years ago
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