Respuesta :
Answer:
E = 1.024 x 10⁴ V/m
Explanation:
given,
KE of the particle = 3.25 x 10⁻¹⁵ J
mass of the proton = 1.673 x 10⁻²⁷ Kg
charge of the proton = 1.602 x 10⁻¹⁹ C
distance = 2 m
Electric field = ?
we know
work done = KE, and also
W = q V
now,
[tex]V = \dfrac{W}{q}[/tex]
[tex]V = \dfrac{3.25\times 10^{-15}}{1.602\times 10^{-19}}[/tex]
V = 2.028 x 10⁴ V
now, using equation of electric field
[tex]E = \dfrac{V}{d}[/tex]
[tex]E = \dfrac{2.028\times 10^4}{2}[/tex]
E = 1.024 x 10⁴ V/m
hence, the magnitude of the electric field is equal to E = 1.024 x 10⁴ V/m
Answer:
E = 1.024 x 10⁴ V/m
Explanation:
given,
KE of the particle = 3.25 x 10⁻¹⁵ J
mass of the proton = 1.673 x 10⁻²⁷ Kg
charge of the proton = 1.602 x 10⁻¹⁹ C
distance = 2 m
Electric field = ?
we know
work done = KE, and also
W = q V
now,
V = 2.028 x 10⁴ V
now, using equation of electric field
E = 1.024 x 10⁴ V/m
hence, the magnitude of the electric field is equal to E = 1.024 x 10⁴ V/m
