Answer :
Answer:
20 m/s
Explanation:
The force experienced by a charged particle in an electric field is given by
[tex]F=qE[/tex]
where, in this problem:
[tex]q=80 mC=0.080 C[/tex] is the charge of the particle
E is the electric field
The electric field here has components:
[tex]E_x=-2.5 N/C\\E_y=0\\E_z=0[/tex]
So the components of the force experienced by the particle are:
[tex]F_x=qE_x=(0.080)(-2.5)=-0.2 N\\F_y=0\\F_z=0[/tex]
Now we can find the components of the acceleration experienced by the particle, using Newton's second law of motion:
[tex]a=\frac{F}{m}[/tex]
where
m = 4.0 g = 0.004 kg is the mass of the particle
The 3 components of the acceleration are:
[tex]a_x=\frac{F_x}{m}=\frac{-0.2}{0.004}=-50 m/s^2\\a_y=0\\a_z=0[/tex]
Now we can find the components of the velocity of the particle at time t using the suvat equation:
[tex]v=u+at[/tex]
where:
[tex]u_x=80 m/s\\u_y=0\\u_z=0[/tex]
are the initial components of the velocity
Therefore, at t = 2.0 s, we have:
[tex]v_x=u_x+a_xt=80+(-50)(2.0)=-20 m/s\\v_y=u_y+a_yt=0+0=0\\v_z=u_z+a_zt=0+0=0[/tex]
And so, the speed of the particle is the magnitude of the final velocity:
[tex]v=\sqrt{v_x^2+v_y^2+v_z^2}=\sqrt{(-20)^2+0+0}=20 m/s[/tex]
The speed of particle at t = 2.0 s would be:
20 m/s
Electric field
According to the question,
Mass, m = 4.0 g
Charge, q = 80 mC
Uniform electric field, Ex = -2.5 N/C
Velocity, [tex]v_x[/tex] = 80 m/s
[tex]v_y[/tex] = [tex]v_z[/tex] = 0
Time, t = 2.0 s
Now,
The force on the particle will be:
→ F = qE
By substituting the values,
= 0.080 × -2.5
= -0.200 Newtons
then,
→ a = [tex]\frac{F}{m}[/tex]
= [tex]-\frac{0.200}{0.004}[/tex]
= -50 m/s²
hence,
The final speed be:
= Initial speed + at
= 80 + (-50) × 2.0
= -20 m/s
Thus the above answer is correct.
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