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6. An α particle is a positively charge particle which is the nucleus of a helium atom He. An engineer who studied
Physics 158 E&M is designing a detector using a long metal wire and a uniform plastic plate. The details are as follows
-- the long line carrying a uniform linear charge density +50.0 μC/m runs parallel to and 10.0cm from the surface of a
large, flat plastic plate that has a uniform surface charge density of −100μC/m2 on one side.
a) Determine the individual Electric Fields of the wire and the plate.
b) Determine the total Electric Field of the wire and plate.
c) Determine the location of all points where an α particle would feel no force due to this arrangement of
charged objects. Assume the plastic plate lies in the (x,y) plane.

Look at the attached image please.

6 An α particle is a positively charge particle which is the nucleus of a helium atom He An engineer who studied Physics 158 EampM is designing a detector using class=

Respuesta :

The electric field due to the line and plate can be obtained by plugging

the values of +50μC/m and -100 μC/m² into the respective formulas.

Responses:

[tex]a) \hspace{0.15 cm} Electric \ field \ of \ the \ line, \ E_z \approx \underline{\dfrac{899180.469}{z}}[/tex]

[tex]Electric \ field \ from \ the \ plate, \ E_p = \underline{ 5649717.51 \ N/C}[/tex]

[tex]b) \hspace{0.15 cm}Total \ electric \ field, \ E_{Total} = \underline{\left(\dfrac{899180.469}{z}+ 5649717.51 \right) \ N/C}[/tex]

c) The alpha particle will feel no force at a distance 15.9 cm from the line in the z-plane

Which method can be used to calculate the electric fields in the system?

The electric field of the line is given as follows;

  • [tex]E_z = \mathbf{\dfrac{\lambda}{2 \cdot \pi \cdot \epsilon _0 \cdot z}}[/tex]

Where;

λ = +50.0 μC/m

ε₀ = The permittivity of free space = 8.85 × 10⁻¹² F/m

z = The distance from the plate

Therefore;

  • [tex]Electric \ field \ of \ the \ line, \ E_z = \dfrac{+50 \times 10^{-6}}{2 \times \pi \times 8.85 \times 10^{-12} \cdot z} \approx \underline{\dfrac{899180.469}{z}}[/tex]

The electric field due to the plate is given as follows;

[tex]\mathbf{Potential} \ at \ a \ distance \ x, \ V = -\dfrac{\sigma \cdot x }{2 \cdot \epsilon _0}[/tex]

Which gives;

[tex]Electric \ field \ from \ the \ plate, \ E_p= \mathbf{-\dfrac{\sigma }{2 \cdot \epsilon _0}}[/tex]

Therefore;

  • [tex]Electric \ field \ from \ the \ plate, \ E_p= -\dfrac{ -100 \times 10^{-6} \ C/m^2 }{2 \times 8.85 \times 10^{-12} \, C^2/(N \cdot m^2)} = \underline{ 5649717.51 \ N/C}[/tex]

b) The total electric field of the wire and the plate is therefore;

[tex]E_{Total} = \mathbf{ E_z + E_p}[/tex]

Which gives;

  • [tex]Total \ electric \ field, \ E_{Total} = \underline{\left(\dfrac{899180.469}{z}+ 5649717.51 \right) \ N/C}[/tex]

c) The points at which an alpha particle will fill no force is found as follows;

F = Q·E

The charge on an alpha particle, Q = 3.2 × 10⁻¹⁹ C

At the point the alpha particle feels no charge, we have;

[tex]\dfrac{899180.469 \times 3.2 \times ^{-19}}{z}= \mathbf{5649717.51 \right) \times 3.2 \times ^{-19}}[/tex]

Which gives;

[tex]z = \dfrac{899180.469 \times 3.2 \times ^{-19}}{5649717.51 \times 3.2 \times ^{-19}} \approx \mathbf{ 0.159}[/tex]

The distance from the long line at which an alpha particle will feel no force is z ≈ 0.159 m = 15.9 cm Which gives;

  • A radius of 15.9 cm in the z-plane

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