Answer:
The magnitude of the electric field is 512171146711.7046 N/C.
The direction is toward the electron as electron has negative charge.
Explanation:
e = Charge of electron = [tex]1.6\times 10^{-19}\ C[/tex]
r = Radius of circular orbit = [tex]5.3\times 10^{-11}\ m/s[/tex]
[tex]\epsilon_0[/tex] = Permittivity of free space = [tex]8.85\times 10^{-12}\ F/m[/tex]
The Electric field is given by
[tex]E=\frac{e}{4\pi\epsilon_0 r^2}\\\Rightarrow B=\frac{1.6\times 10^{-19}}{4\pi\times 8.85\times 10^{-12}\times (5.3\times 10^{-11})^2}\\\Rightarrow E=512171146711.7046\ N/C[/tex]
The magnitude of the electric field is 512171146711.7046 N/C.
The direction is toward the electron as electron has negative charge.
The direction of electric field is 512171146711.7046 N/C.
From the question,
e = Charge of electron is 1.6× 10-9
r = Radius of circular orbit is 5.3× 10-11
Permittivity of free space is 8.85 × 10-12
The Electric field is written below.
E=c/4πeor^2
E= 1.6×10-19/4π × 8.85×5.3×10-11
E= 512171146711.7N/c.
Electric field refer to field that have charged particles both positive and negative charge.
Therefore, Therefore, The direction of electric field is 512171146711.7046 N/C.
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