Chapter 21, Problem 61A

(a)

To determine

The direction of the force on a negatively charged particle.

Answer to Problem 61A

Upwards

Explanation of Solution

Given:

Electric field strength, $E=150\text{N/C}$ , downwards

Formula Used:

Electric field strength can be obtained by:

$E=\frac{F}{q}$

Here, F is the electrostatic force and q is the test charge.

The direction of electric field is along the direction of the force for the positive test charge and opposite for the negative test charge.

Calculation:

For a negatively charge particle, the direction of the electrostatic force would be opposite the direction of the electric field. Thus, it will be upwards.

Conclusion:

The direction of force on a negatively charged particle would be opposite to the direction of the electric field.

(b)

To determine

The electric force on an electron in the given field.

Answer to Problem 61A

$2.4\times {10}^{-17}\text{N}$ in the upward direction.

Explanation of Solution

Given:

Electric field strength, $E=150\text{N/C}$ , downwards

Charge on an electron, $q=-1.6\times {10}^{-19}\text{C}$

Formula Used:

Electrostaticforceon a charge particle can be obtained by:

$F=qE$

Here, E is the strength of theelectricfield and q is the test charge.

The direction of electric field is along the direction of the force for the positive test charge and opposite for the negative test charge.

Calculation:

Substitute the values and solve:

$\begin{array}{c}F=qE\\ =\left(-1.6\times {10}^{-19}\text{C}\right)\left(150\text{N/C}\right)\\ =-2.4\times {10}^{-17}\text{N}\end{array}$

Conclusion:

The electric force on an electron is $2.4\times {10}^{-17}\text{N}$ in the upward direction.

(c)

To determine

To Compare: The strength of electric force on an electron with the force of gravity on it.

Explanation of Solution

Given:

The electric force on an electron is $2.4\times {10}^{-17}\text{N}$ in the upward direction.

Mass of an electron, $m_e=9.1\times {10}^{-31}\text{kg}$

Formula Used:

The force of gravity on the electron can be obtained by:

$F_g=m_{e}g$

Here, g is the acceleration due to gravity.

Calculation:

Substitute the values and solve:

$\begin{array}{c}{F}_g=m_{e}g\\ =\left(9.1\times {10}^{-31}\text{kg}\right)\left(9.81\text{N/kg}\right)\\ =8.9\times {10}^{-30}\text{N}\end{array}$

$\begin{array}{c}\frac{F_e}{F_g}=\frac{2.4\times {10}^{-17}\text{N}}{8.9\times {10}^{-30}\text{N}}\\ =2.7\times {10}^{12}\end{array}$

Conclusion:

The force of gravity is very less than the electric force on an electron.