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

The star is moving away from our planet. To elaborate on the Doppler shift: This phenomenon, related to the Doppler effect, is the variation in the perceived frequency or wavelength (color) of a wave when the source of the waves and the observer are in motion relative to one another. Consequently, it can be inferred that as an object recedes, it exhibits more redshift in its spectrum. For instance, when a star moves away, its spectral lines shift towards the red end of the spectrum, whereas if it approaches Earth, the spectral lines move towards blue. Given that the peak wavelength is roughly 650 nm—which is associated with red—it can be concluded that the star is indeed moving away from Earth.

Answer:

Part A) Electric fields at the designated point due to charges q₁ and q₂:

E₁ = 33.75 * 10³ N/C (-j), E₂ = (6.48 (-i) + 8.64 (+j)) * 10³ N/C

Part B) The overall electric field at P (Ep)

Ep = (6.48 * 10³ (-i) + 25.11 * 10³ (-j)) N/C

Explanation:

Conceptual analysis

The electric field at point P caused by a point charge is calculated as:

E = k*q/d²

E: Electric field measured in N/C

q: charge magnitude in Newtons (N)

k: electric constant measured in N*m²/C²

d: distance from the charge q to point P in meters (m)

Equivalence:

1 nC = 10⁻⁹ C

1 cm = 10⁻² m

Data:

k = 9 * 10⁹ N*m²/C²

q₁ = -6.00 nC = -6 * 10⁻⁹ C

q₂ = +3.00 nC = +3 * 10⁻⁹ C

d₁ = 4 cm = 4 * 10⁻² m

d₂ = 5 * 10⁻² m

Part A) Calculation for electric fields at point from q₁ and q₂:

Refer to the attached illustration:

E₁: Electric Field at point P(0,4) cm due to charge q₁. Since q₁ is negative (q₁-), the electric field approaches the charge.

E₂: Electric Field at point P(0,4) cm due to charge q₂. Since q₂ is positive (q₂+), the electric field emanates from the charge.

E₁ = k*q₁/d₁² = 9 * 10⁹ * 6 * 10⁻⁹ / (4 * 10⁻²)² = 33.75 * 10³ N/C

E₂ = k*q₂/d₂²= 9 * 10⁹ * 3 * 10⁻⁹ / (5 * 10⁻²)² = 10.8 * 10³ N/C

E₁ = 33.75 * 10³ N/C (-j)

E₂x = E₂cosβ = 10.8 * (3/5) = 6.48 * 10³ N/C

E₂y = E₂sinβ = 10.8 * (4/5) = 8.64 * 10³ N/C

E₂ = (6.48 (-i) + 8.64 (+j)) * 10³ N/C

Part B) Calculation for net electric field at P (Ep)

The electric field at point P from multiple point charges is the vector sum of the individual electric fields.

Ep = Epx (i) + Epy (j)

Epx = E₂x = 6.48 * 10³ N/C (-i)

Epy = E₁y + E₂y = (33.75 * 10³ (-j) + 8.64 * 10³ (+j)) N/C = 25.11 * 10³ (-j) N/C

Ep = (6.48 * 10³ (-i) + 25.11 * 10³ (-j)) N/C

Ep = (6.48 * 10³ (-i) + 25.11 * 10³ (-j)) N/C

It equals 5z

Explanation:

5z

Response:

 y_red / y_blue = 1.11

Clarification:

To determine the image for each wavelength, we'll utilize the lens maker's equation

         1 /f = 1 /o + 1 /i

Where f signifies the focal length, o represents the object distance, and i indicates the image distance

For red light

           1 / i = 1 / f - 1 / o

           1 / i_red = 1 / f_red - 1 / o

           1 / i_red = 1 / 19.57 - 1/30

           1 / i_red = 1.776 10-2

           i_red = 56.29 cm

For blue light

            1 / i_blue = 1 / f_blue - 1 / o

            1 / i_blue = 1 / 18.87 - 1/30

            1 / i_blue = 1.966 10-2

            i_blue = 50.863 cm

Next, we will compute the magnification ratio

             m = y ’/ h = - i / o

             y ’= - h i / o

For red light

            y_red ’= - 5 56.29 / 30

            y_red ’= - 9.3816 cm

For blue light

            y_blue ’= 5 50.863 / 30

            y_blue ’= - 8.47716 cm

The ratio of the heights of both images is

            y_red ’/ y_blue’ = 9.3816 / 8.47716

            y_red / y_blue = 1.107

            y_red / y_blue = 1.11

It's not only the horizontal segments of the loop that extend into the magnetic field. We can ignore the horizontal sections that dip into the magnetic fields since only the sections that are perpendicular to the magnetic field have a significant contribution to it.

<span> </span>