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Q54PE

Expert-verifiedFound in: Page 666

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**Earth has a net charge that produces an electric field of approximately **\(150{\rm{ N/C}}\)** downward at its surface. (a) What is the magnitude and sign of the excess charge, noting the electric field of a conducting sphere is equivalent to a point charge at its center? (b) What acceleration will the field produce on a free electron near Earth’s surface? (c) What mass object with a single extra electron will have its weight supported by this field?**

(a) The amount of the extra charge is \( - 6.78 \times {10^5}{\rm{ C}}\).

(b) The acceleration of the free electron is \(2.64 \times {10^{13}}{\rm{ m/}}{{\rm{s}}^{\rm{2}}}\) and is directed upward.

(c) The mass of the object is \(2.45 \times {10^{ - 18}}{\rm{ kg}}\).

**The region in which a charge can show its effect is known as electric field. When another charge comes in this region, it will experience some electrostatic force.**

The expression for the electric field is,

\(E = \frac{{Kq}}{{{r^2}}}\)

Here, \(K\) is the electrostatic force constant, \(q\) is the charge, and \(r\) is the distance of the point of consideration for the charge.

The extra charge can be calculated using equation (1.1).

Rearranging equation (1.1) in order to get an expression for the charge.

\(q = \frac{{E{r^2}}}{K}\)

Here, \(E\) is the electric field (\(E = - 150{\rm{ N/C}}\), here negative sign indicates that the electric field is directed downwards), \(r\) is the radius of Earth \(\left( {r = 6.378 \times {{10}^6}{\rm{ m}}} \right)\), and \(K\) is the electrostatic force constant \(\left( {K = 9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)\).

Substituting all known values,

\(\begin{array}{c}q = \frac{{\left( { - 150{\rm{ N/C}}} \right) \times {{\left( {6.378 \times {{10}^6}{\rm{ m}}} \right)}^2}}}{{\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)}}\\ = - 6.78 \times {10^5}{\rm{ C}}\end{array}\)

Hence, the amount of the extra charge is \( - 6.78 \times {10^5}{\rm{ C}}\).

The force on a free electron when placed in an electric field is,

\(F = qE\)

Here, \(q\) is the charge on electron \(\left( {q = - 1.6 \times {{10}^{ - 19}}{\rm{ }}C} \right)\), and \(E\) is the electric field (\(E = - 150{\rm{ N/C}}\), here negative sign indicates that the electric field is directed downwards).

The force on the electron is,

\(F = ma\)

Here, \(m\) is the mass of the electron \(\left( {m = 9.1 \times {{10}^{ - 31}}{\rm{ kg}}} \right)\), and \(a\) is the acceleration of the free electron.

Equating equations (1.2) and (1.3),

\(ma = qE\)

Rearranging the equation in order to get an expression for the acceleration of the free electron.

\(a = \frac{{qE}}{m}\)

Substituting all known values,

\(\begin{array}{c}a = \frac{{\left( { - 1.6 \times {{10}^{ - 19}}{\rm{ C}}} \right) \times \left( { - 150{\rm{ N/C}}} \right)}}{{\left( {9.1 \times {{10}^{ - 31}}{\rm{ kg}}} \right)}}\\ = 2.64 \times {10^{13}}{\rm{ m/}}{{\rm{s}}^{\rm{2}}}\end{array}\)

Since, the acceleration and electric field is in opposite direction. The electron will accelerate upwards.

Hence, the acceleration of the free electron is \(2.64 \times {10^{13}}{\rm{ m/}}{{\rm{s}}^{\rm{2}}}\) and is directed upward.

The weight of the object is,

\({F_g} = mg\)

Here, \(m\) is the mass of the object, and \(g\) is the acceleration due to gravity \(\left( {g = 9.81{\rm{ }}m/{s^2}} \right)\).

Since, the weight of the object is supported by the electrostatic force due to the electric field. Thus,

\(\begin{array}{c}{F_g} = F\\mg = qE\end{array}\)

Rearranging the equation in order to get an expression for the mass of the object.

\(m = \frac{{qE}}{g}\)

Substituting all known values,

\(\begin{array}{c}m = \frac{{\left( { - 1.6 \times {{10}^{ - 19}}{\rm{ C}}} \right) \times \left( { - 150{\rm{ N/C}}} \right)}}{{\left( {9.81{\rm{ m/}}{{\rm{s}}^{\rm{2}}}} \right)}}\\ = 2.45 \times {10^{ - 18}}{\rm{ kg}}\end{array}\)

Hence, the mass of the object is \(2.45 \times {10^{ - 18}}{\rm{ kg}}\).

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