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Expert-verified Found in: Page 666 ### College Physics (Urone)

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 $$2.45 \times {10^{ - 18}}{\rm{ kg}}$$.

See the step by step solution

## Step 1: Electric field

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.

## Step 2: Magnitude and sign of extra 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 $$- 6.78 \times {10^5}{\rm{ C}}$$.

## Step 3: Acceleration of the free electron

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.

## Step 4: (c) Mass of the object

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 $$2.45 \times {10^{ - 18}}{\rm{ kg}}$$. ### Want to see more solutions like these? 