Question

A proton \(\left( {m = 1.67 \times 1{0^{ - 27}} kg} \right)\) is suspended at rest in a uniform electric field \(\overrightarrow {\bf{E}} \). Take into account gravity at the Earth’s surface, and determine \(\overrightarrow {\bf{E}} \).

Short answer

The magnitude of the electric field is \(1.02 \times {10^{ - 7}}{\rm{ N/C}}\). The direction of the electric field is upward.

Step-by-step solution

Understanding electric field vector

The magnitude of the electric field vector can be obtained with the help of the electric force vector and the value of the charge. The direction of the electric field is the same as the direction of the force.

The electric force acting on the proton equals the gravitational force acting on it.

Identification of given data

The given data can be listed below as:

• The charge on the proton is\(q = 1.6 \times {10^{ - 19}}{\rm{ C}}\).
• The mass of the proton is\(m = 1.67 \times {10^{ - 27}}{\rm{ kg}}\).
• The acceleration due to gravity is \(g = 9.81{\rm{ m/}}{{\rm{s}}^2}\).

Determination of the magnitude and direction of the electric field

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

\({F_e} = qE\) … (i)

Here, Eis the magnitude of the electric field.

The magnitude of the force due to its weight can be expressed as:

\({F_g} = mg\) … (ii)

Here, gis the acceleration due to gravity.

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

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

Substitute the values in the above equation.

\(\begin{aligned}{c}E &= \frac{{1.67 \times {{10}^{ - 27}}{\rm{ kg}} \times 9.81{\rm{ m/}}{{\rm{s}}^2} \times \left( {\frac{{1{\rm{ N}}}}{{1{\rm{ kg}} \cdot {\rm{m/}}{{\rm{s}}^2}}}} \right)}}{{1.6 \times {{10}^{ - 19}}{\rm{ C}}}}\\ &= \frac{{1.64 \times {{10}^{ - 26}}{\rm{ N}}}}{{1.6 \times {{10}^{ - 19}}{\rm{ C}}}}{\rm{ }}\\ &= 1.02 \times {10^{ - 7}}{\rm{ N/C}}\end{aligned}\)

The charge of a proton is positive. The acceleration due to gravity acts in the downward direction. The electric field will be in the opposite direction of the acceleration. The electric force should be in the upward direction. The direction of the electric field will be along the direction of the electric force.

Thus, the magnitude of the electric field is \(1.02 \times {10^{ - 7}}{\rm{ N/C}}\) and it is directed upward.