Question

A proton is observed to have an instantaneous acceleration of 10 × 1011 m/s2. What is the magnitude of the electric field at the proton's location?

Answer 1

To solve this problem we will apply the concept of Newton's second law with which we will obtain the strength of the proton. We know the mass and the acceleration is given in the statement. Subsequently said Force by equilibrium can be matched the electrostatic force of Coulomb, defined as the product between the charge and the electric field. Our values are

$m = 1.67*10^{-27}kg$

$a = 10*10^{11} m/s^2$

Applying the Newton's second law,

$F = ma$

$F = (1.67*10^{-27}kg)( 10*10^{11} m/s^2)$

$F = 1.67*10^{-15}N$

By the Coulomb's equation for electrostatic Force we have that

$F = qE$

Remember that the charge of a proton is $1.6*10^{-19}C$

Replacing we have,

$1.67*10^{-15} = 1.6*10^{-19} E$

$E = 10437.5 N/C$

Therefore the magnitude of the electric field at the proton's location is $10437.5 N/C$

Answer 2

The magnitude of the electric field at the proton's location is 10,437.5 N/C.

The given parameters:

• Acceleration of the proton, a = 10 x 10¹¹ m/s²
• Mass of proton, m = 1.67 x 10⁻²⁷ kg

The magnitude of the electric field at the proton's location is calculated as follows;

F = ma

F = qE

qE = ma

$E = \frac{ma}{q} \\\\E = \frac{1.67 \times 10^{-27} \times 10\times 10^{11}}{1.6 \times 10^{-19}} \\\\E = 10,437.5 \ N/C$

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