Question

A proton starts from rest and gains 8.35 x 10^-14 joule of kinetic energy as it accelerates between points A and B in an electric field. The potential difference between points A and B in the electric field is approximately

Answer 1

Answer:

5.22 x 10^5 V

Explanation:

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Answer 2

The potential difference between points A and B in the electric field is approximately equal to $5.22 \times 10^5\;V$

Given the following data:

• Kinetic energy = $8.35 \times 10^{-14} \;Joule$

Scientific data:

• Charge of proton = $1.6 \times 10^{-19}\;C$

To determine the potential difference between points A and B in the electric field:

Mathematically, kinetic energy is calculated by using this formula;

$K.E = qV_d$

Where:

• K.E is the kinetic energy.

• q is the charge of a particle.

• $V_d$ is the potential difference of a particle.

Making $V_d$ the subject of formula, we have:

$V_d = \frac{K.E}{q}$

Substituting the given parameters into the formula, we have;

$V_d = \frac{8.35 \times 10^{-14} }{1.6 \times 10^{-19}}\\\\V_d = 521875$

When $V_d$ is approximated, we have:

Potential difference = 522,000 Volts

Potential difference = $5.22 \times 10^5\;Volts$

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