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Estimate the magnitude of the electric field due to the proton in a hydrogen atom at a distance of 5.29 x 10-11 m, the expected position of the electron in the atom.
a) 1012 N/C
b) 10-11 NIC
c) 106 N/C
d) 1014 N/C
e) 108 N/C

Respuesta :

carlycundiff2007
Would be A 1012 N/C because The magnitude of the electric field at distance r from a point charge q is E=k
e
​
q/r
2
, so
E=
(5.11×10
−11
m)
2

(8.99×10
9
N.m
2
/C
2
)(1.60×10
−19
C)
​

=5.51×10
11
N/C∼10
1
2N/C
making (e) the best choice for this question.

The magnitude of the electric field due to the proton in a hydrogen atom at a distance of 5.29 x 10-11 m, the expected position of the electron in the atom will be a) 1012 N/C

What is electric field due to a charge ?

Electric field is defined as the electric force per unit charge. The direction of the field is taken to be the direction of the force it would exert on a positive test charge.

electric field = k * q / r^2

given

distance (r) =  5.29 x 10-11 m

q of proton = + 1.602 * 10 ^ -19 C

k = 9 * 10 ^9 Nm^2 / C^2

Electric field = (9 * 10 ^9 Nm^2 / C^2 *  1.602 * 10 ^ -19 C )/  (5.29 x 10-11 m)^2

                    = 14.418 * 10 ^ (-10) / 27.98 * 10 ^ (-22)

                   = 0.5153 * 10 ^12 N/C

correct option a) 1012 N/C

learn more about Electric field

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