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A particle has a charge of +1.5 µC and moves from point A to point B, a distance of 0.15 m. The particle experiences a constant electric force, and its motion is along the line of action of the force. The difference between the particle's electric potential energy at A and B is EPEA - EPEB = +9.0 10-4 J.
(a) Find the magnitude and direction of the electric force that acts on the particle. _______N the direction of motion (along, against, perpendicular)
(b) Find the magnitude and direction of the electric field that the particle experiences. _______N/C the direction of motion (along, against, perpendicular)

Respuesta :

Answer:

Part a)

[tex]F = 6 \times 10^{-3} N[/tex]

Direction of force is along the motion of charge

Part b)

[tex]E = 4000 N/C[/tex]

direction of electric field is along the direction of motion

Explanation:

Part a)

As we know that the change in electric potential energy is equal to the work done by electric field

[tex]W = EPE_A - EPE_B[/tex]

[tex]W = 9.0 \times 10^{-4} J[/tex]

now from the equation of work done we know that

[tex]W = F.d[/tex]

[tex](9.0 \times 10^{-4}) = F(0.15)[/tex]

[tex]F = 6 \times 10^{-3} N[/tex]

Direction of force is along the motion of charge

Part b)

As we know the relation between electrostatic force and electric field given as

[tex]F = qE[/tex]

[tex](6 \times 10^{-3}) = 1.5 \times 10^{-6} E[/tex]

[tex]E = 4000 N/C[/tex]

direction of electric field is along the direction of motion

The magnitude of the electric field is 4000 N/C while the direction of the electric field is along the direction of the motion of the charge.

Given to us

Charge of the particle, q = +1.5 µC = 1.5 x 10⁻⁶

Distance between points A and B, d = 0.15 m

The particle's electric potential energy at A and B

= [tex]\rm EPE_A - EPE_B[/tex] = +9.0x10⁻⁴ J

What is the magnitude and direction of the electric force that acts on the particle?

The electric charge in electric potential energy is equal to the work done by an electric field. therefore,

[tex]\rm W=EPE_A - EPE_B[/tex]

W = 9.0x10⁻⁴ J

We know that work done by the system can be given as,

[tex]\rm Work = force\times displacement[/tex]

Substitute the values,

[tex]9 \times 10^{-4}= F \times 0.15[/tex]

F = 1.5 x 10⁻³ N

Thus, the magnitude of the force is 1.5 x 10⁻³ N while its direction is along with the motion of the charge.

What is the magnitude and direction of the electric field that the particle experiences?

We know the relationship between electrostatic and electric force, therefore,

[tex]\rm Force = charge\times Electric\ field[/tex]

Substitute the values,

[tex]6\times 10^{-3}=1.5\times 10^{-6} \times \rm E[/tex]

E = 4000 N/C

Hence, the magnitude of the electric field is 4000 N/C while the direction of the electric field is along the direction of the motion of the charge.

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