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Elis [28]
2 years ago
12

A proton moves a distance 10 cm in a uniform electric field of 3.5 kN C, in the direction of the field.

Physics
1 answer:
lawyer [7]2 years ago
7 0

The change in potential energy of the proton is  5.6 x 10^{-17} Joule

<h3>What is a Uniform Electric Field ?</h3>

A uniform electric field is where the electric field strength is the same at all points in the field. In the uniform field, the force experienced by a charge is the same no matter where it is placed in the field.

Given that a proton moves a distance 10 cm in a uniform electric field of 3.5 kN C, in the direction of the field.

  • The distance d = 10 cm = 0.1 m
  • Electric field E = 3.5 KN/C
  • Proton charge q = 1.6 x 10^{-19} C

The Work done = Fd

but F = Eq

Recall that Electric field E = F/q = V/d

Where V = potential difference.

Let us first calculate the V

E = V/d

V = Ed

Substitute all the parameters into the formula above

V = 3.5 × 10³ × 0.1

V = 350 v

from F/q = V/d

make F the subject of formula and substitute it in work formula

F = Vq/d

W.D = Vq/d x d

W.D = Vq

Substitute all the parameters into the formula above

W.D = 350 x 1.6 x 10^{-19}

W.D = 5.6 x 10^{-17} J

Work done = Energy = Potential Energy

Therefore, the change in potential energy of the proton is 5.6 x 10^{-17}<em> Joule</em>

<em />

Learn more about Electric Field here: brainly.com/question/14372859

#SPJ1

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An electron moves at 0.130 c as shown in the figure (Figure 1). There are points: A, B, C, and D 2.10 μm from the electron.
Olegator [25]

Hi there!

We can use Biot-Savart's Law for a moving particle:
B= \frac{\mu_0 }{4\pi}\frac{q\vec{v}\times \vec{r}}{r^2 }

B = Magnetic field strength (T)
v = velocity of electron (0.130c = 3.9 × 10⁷ m/s)

q = charge of particle (1.6 × 10⁻¹⁹ C)

μ₀ = Permeability of free space (4π × 10⁻⁷ Tm/A)

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There is a cross product between the velocity vector and the radius vector (not a quantity, but specifies a direction). We can write this as:

B= \frac{\mu_0 }{4\pi}\frac{q\vec{v} \vec{r}sin\theta}{r^2 }

Where 'θ' is the angle between the velocity and radius vectors.

a)
To find the angle between the velocity and radius vector, we find the complementary angle:

θ = 90° - 60° = 30°

Plugging 'θ' into the equation along with our other values:

B= \frac{\mu_0 }{4\pi}\frac{q\vec{v} \vec{r}sin\theta}{r^2 }\\\\B= \frac{(4\pi *10^{-7})}{4\pi}\frac{(1.6*10^{-19})(3.9*10^{7}) \vec{r}sin(30)}{(2.1*10^{-5})^2 }

B = \boxed{7.07 *10^{-10} T}

b)
Repeat the same process. The angle between the velocity and radius vector is 150°, and its sine value is the same as that of sin(30°). So, the particle's produced field will be the same as that of part A.

c)

In this instance, the radius vector and the velocity vector are perpendicular so

'θ' = 90°.

B= \frac{(4\pi *10^{-7})}{4\pi}\frac{(1.6*10^{-19})(3.9*10^{7}) \vec{r}sin(90)}{(2.1*10^{-5})^2 } = \boxed{1.415 * 10^{-9}T}

d)
This point is ALONG the velocity vector, so there is no magnetic field produced at this point.

Aka, the radius and velocity vectors are parallel, and since sin(0) = 0, there is no magnetic field at this point.

\boxed{B = 0 T}

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Answer:

Check explanation

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