Respuesta :
Answer:
(a) Work done will be [tex]2\times 10^{-5}[/tex]
(b) kinetic energy of the charge particle is [tex]Ke=2\times 10^{-5}[/tex]
Explanation:
We have given the charge carried by the particle [tex]q=4nC=4\times 10^{-9}C[/tex]
Electric field [tex]E=2\times 10^3N/C[/tex]
Distance through which particle move d = 2.5 m
We know that electric force is given by
[tex]F=qE=4\times 10^{-9}\times 2\times 10^3=8\times 10^{-6}N[/tex]
Now we know that work done is given by
[tex]W=Fd=8\times 10^{-6}\times 2.5=20\times 10^{-6}j=2\times 10^{-5}J[/tex]
(B) We know that work done is given by change in kinetic energy
So [tex]W=\frac{1}{2}mv_1^2-\frac{1}{2}mv_2^2[/tex]
[tex]2\times 10^{-5}=Ke-0[/tex]
[tex]Ke=2\times 10^{-5}[/tex]
So kinetic energy of the charge particle is [tex]Ke=2\times 10^{-5}[/tex]
The electrostatic work done on the particle after it has moved the given distance is 2 x 10⁻⁵ J.
The kinetic energy of the particle at the given instant is 2 x 10⁻⁵ J.
The given parameters;
- charge on the particle, Q = 4 nC
- magnitude of the magnetic field, E = 2 x 10³ N/C
- distance moved by the particle, d = 2.5 m
The electrostatic work done on the particle after it has moved the given distance is calculated as follows;
W = Fd
W = (EQ) x d
W = (2 x 10³ x 4 x 10⁻⁹) x 2.5
W = 2 x 10⁻⁵ J
Based on work energy theorem, the kinetic energy of the particle at the given instant is equal to the work done on the particle at that time.
Kinetic energy = work done
Kinetic energy = 2 x 10⁻⁵ J
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