Respuesta :
Answer:
The magnitude of charge on the particle is [tex]1.6\times 10^{-19}\ C[/tex].
Explanation:
Given that,
The magnitude of electric field, E = 215 N/C
The electrical potential energy of the charge decreases by, [tex]-6.9\times 10^{-19}\ J[/tex] as it moves.
We need to find the magnitude of the charge on the moving particle. The change in electric potential energy is given by :
[tex]\Delta U=q\Delta V\\\\q=\dfrac{\Delta U}{\Delta V}[/tex].........(1)
The electric potential in terms of electric field is given by :
[tex]\Delta V=-Ed\\\\\Delta V=-215\times 0.02=-4.3\ V[/tex]
Equation (1) becomes :
[tex]q=\dfrac{-6.9\times 10^{-19}}{-4.3}\\\\q=1.6\times 10^{-19}\ C[/tex]
So, the magnitude of charge on the particle is [tex]1.6\times 10^{-19}\ C[/tex].
The magnitude of charge on the moving particle is [tex]1.6*10^{-19}C[/tex].
Electric Potential :
Given that,
Change in electrical potential energy,[tex]\Delta U=-6.9*10^{-19} J[/tex]
Electric field [tex]E=215N/C[/tex] , Distance [tex]d=2cm=0.02m[/tex]
Change in electric potential is given as,
[tex]\Delta V=-Ed\\\\\Delta V = -215*0.02=-4.3V[/tex]
the magnitude of the charge on the moving particle is,
[tex]q=\frac{\Delta U}{\Delta V} =\frac{-6.9*10^{-19} }{-4.3} \\\\q=1.6*10^{-19}C[/tex]
The magnitude of charge on the moving particle is [tex]1.6*10^{-19}C[/tex].
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