Question

(II) What is the electric field strength at a point in space where a proton experiences an acceleration of 2.4 million “g’s”?

Short answer

The magnitude of the electric field strength of a proton is \(0.25{\rm{ N/C}}\).

Step-by-step solution

Understanding the electric field strength

It is the strength or power of any electric field at a specific point.In the question, the electric field strength depends on the proton’s mass, acceleration and charge. It is measured in terms of Newton per Coulomb.

Identification of given data

The given data is listed below:

• The acceleration due to gravity is\(g = 9.81{\rm{ m/}}{{\rm{s}}^2}\).
• The acceleration of a proton is\(a = 2.4{\rm{ million}}\left( g \right) = 2.4{\rm{ million}}\left( {\frac{{{{10}^6}}}{{1{\rm{ million}}}}} \right) \times 9.81{\rm{ m/}}{{\rm{s}}^2} = 2.35 \times {10^7}{\rm{ m/}}{{\rm{s}}^2}\).
• The charge on the proton is\(q = 1.6 \times {10^{ - 19}}{\rm{ C}}\).
• The mass of a proton is \(m = 1.67 \times {10^{ - 27}}{\rm{ kg}}\).

Determination of the magnitude of the electric field strength

The magnitude of the electric force on a proton can be expressed as follows:

\(F = qE\) … (i)

Here, E is the electric field strength.

From Newton’s second law, the magnitude of the force can be expressed as follows:

\(F = ma\) … (ii)

Here, ais the acceleration of the proton.

Equate equations (i) and (ii). Then, evaluate the expression of the electric field strength.

\(\begin{aligned}{c}qE = ma\\E = \frac{{ma}}{q}\end{aligned}\)

Substitute the values in the above equation.

\(\begin{aligned}{c}E = \frac{{1.67 \times {{10}^{ - 27}}{\rm{ kg}} \times 2.35 \times {{10}^7}{\rm{ m/}}{{\rm{s}}^2}}}{{1.6 \times {{10}^{ - 19}}{\rm{ C}}}}\\ = \frac{{3.92 \times {{10}^{ - 20}}{\rm{ kg}} \cdot {\rm{m/}}{{\rm{s}}^2}}}{{1.6 \times {{10}^{ - 19}}{\rm{ C}}}}\left( {\frac{{1{\rm{ N}}}}{{1{\rm{ kg}} \cdot {\rm{m/}}{{\rm{s}}^2}}}} \right)\\ = 0.25{\rm{ N/C}}\end{aligned}\)

Thus, the magnitude of the electric field strength of the proton is\(0.25{\rm{ N/C}}\).