Question
(I) What is the magnitude of the electric force of attraction between an iron nucleus $(q=+26 e)$ and its innermost electron if the distance between them is $1.5 \times 10^{-12} \mathrm{m}$ ?
Step 1
Here |q1 q2| = 26 e^2, k = 8.99×10^9 N·m^2/C^2, e = 1.602×10^−19 C, r = 1.5×10^−12 m. Compute: 26 e^2 = 26(1.602×10^−19)^2 = 6.6726×10^−37 C^2. k·(26 e^2) = 8.99×10^9 × 6.6726×10^−37 = 5.997×10^−27 N·m^2. r^2 = (1.5×10^−12)^2 = 2.25×10^−24 m^2. Show more…
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(I) What is the magnitude of the electric force of attraction between an iron nucleus $( q = + 26 e )$ and its innermost electron if the distance between them is $1.5 \times 10 ^ { - 12 } \mathrm { m } ?$
(I) What is the magnitude of the electric force of attraction between an iron nucleus $(q= + 26e)$ electron if the distance between them is 1.5 $\times$ 10$^{-12}$m?
ELECTRIC CHARGE AND ELECTRIC FIELD
Coulomb's Law
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