Question

A proton and an electron in a hydrogen atom are separated on the average by about 5.3 × 10−11 m. What is the magnitude and direction of the electric field set up by the proton at the position of the electron? The Coulomb constant is 8.99 × 109 N · m2 /C 2 . 1. 5.12068 × 1011 N/C away from the proton 2. 8.19309 × 10−8 N/C toward the proton 3. 27.1396 N/C away from the proton 4. 27.1396 N/C toward the proton 5. 5.12068 × 1011 N/C toward the proton 6. 8.19309 × 10−8 N/C away from the proton

Answer 1

Answer:

1. 5.12068 × 1011 N/C away from the proton

Explanation:

The electric field produced by a single point charge is given by:

$E=k\frac{q}{r^2}$

where

k is the Coulomb's constant

q is the magnitude of the charge

r is the distance from the charge

In this problem, we have:

$q=1.6\cdot 10^{-19}C$ is the charge of the proton

$r=5.3\cdot 10^{-11} m$ is the distance at which we want to calculate the field

$k=8.99\cdot 10^9 Nm^2C^{-2}$ is the Coulomb's constant

Substituting into the formula,

$E=(8.99\cdot 10^9 Nm^2C^{-2})\frac{1.6\cdot 10^{-19}C}{(5.3\cdot 10^{-11}m)^2}=5.12068\cdot 10^{11} N/C$

And the direction of the electric field produced by a positive charge is away from the charge, so the correct answer is

1. 5.12068 × 1011 N/C away from the proton