Question

Part A

What is the strength of an electric field that will balance the weight of a proton?

Part B

What is the direction of an electric field that will balance the weight of a proton?

What is the direction of an electric field that will balance the weight of a proton?

	upward
	downward

Part C

What is the strength of an electric field that will balance the weight of an electron?

Part D

What is the direction of an electric field that will balance the weight of an electron?

What is the direction of an electric field that will balance the weight of an electron?

	upward
	downward

Answer 1

Concepts and reason

The required concepts to solve the problem are weight and electric field.

Initially, find the weight of the proton and then equate the electric force with the weight of the proton to find the electric field that balances the proton weight. Later, find the weight of the electron and finally, equate the electric force with the weight of the electron to find the electric field that balances the electron weight.

Fundamentals

The force with which the Earth attracts an object is called the weight of the body.

$W = mg$

Here, $m$ is the mass of the object and $g$ is the acceleration due to gravity.

The electric force is the force experienced by a charged object when placed in an electric field. It is given by,

$F = qE$

Here, $q$ is the charge of the object and $E$ is the electric field.

(A)

The weight of the proton is,

$W = mg$

Substitute $1.67 \times {10^{ - 27}}\,{\rm{kg}}$ for $m$ and $9.8\,{\rm{m/}}{{\rm{s}}^2}$ for $g$ to find the weight of the proton.

$\begin{array}{c}\\W = \left( {1.67 \times {{10}^{ - 27}}{\rm{kg}}} \right)\left( {9.8\,{\rm{m/}}{{\rm{s}}^2}} \right)\\\\ = 1.64 \times {10^{ - 26}}\,{\rm{N}}\\\end{array}$

The equation for electric force is,

$F = qE$

Equating the equations,

$\begin{array}{c}\\qE = 1.64 \times {10^{ - 26}}\,{\rm{N}}\\\\E = \frac{{1.64 \times {{10}^{ - 26}}\,{\rm{N}}}}{q}\\\end{array}$

Substitute the charge of a proton $1.6 \times {10^{ - 19}}\,{\rm{C}}$ for $q$ .

$\begin{array}{c}\\E = \frac{{1.64 \times {{10}^{ - 26}}\,{\rm{N}}}}{{1.6 \times {{10}^{ - 19}}\,{\rm{C}}}}\\\\ = 1.023 \times {10^{ - 7}}\,{\rm{N/C}}\\\end{array}$

(B)

The charge on the proton is positive. So, the electric field that balances the weight of the proton is,

$E = 1.023 \times {10^{ - 7}}\,{\rm{N/C}}$ .

As the sign of the electric field is positive, the direction of the electric field that balances the weight of the proton is upwards.

(C)

The weight of the electron is,

$W = mg$

Substitute $9.1 \times {10^{ - 31}}\,{\rm{kg}}$ for $m$ and $9.8\,{\rm{m/}}{{\rm{s}}^2}$ for $g$ to find the weight of the electron.

$\begin{array}{c}\\W = \left( {9.1 \times {{10}^{ - 31}}\,{\rm{kg}}} \right)\left( {9.8\,{\rm{m/}}{{\rm{s}}^2}} \right)\\\\ = 8.92 \times {10^{ - 30}}\,{\rm{N}}\\\end{array}$

The equation for electric force is,

$F = qE$

Equating the equations,

$\begin{array}{l}\\qE = 8.92 \times {10^{ - 30}}\,{\rm{N}}\\\\E = \frac{{8.92 \times {{10}^{ - 30}}\,{\rm{N}}}}{q}\\\end{array}$

Substitute the charge of an electron $- 1.6 \times {10^{ - 19}}\,{\rm{C}}$ for $q$ .

$\begin{array}{c}\\E = \frac{{8.92 \times {{10}^{ - 30}}\,{\rm{N}}}}{{ - 1.6 \times {{10}^{ - 19}}\,{\rm{C}}}}\\\\ = - 5.57 \times {10^{ - 11}}\,{\rm{N/C}}\\\end{array}$

(D)

The charge on the electron is negative. So, the electric field that balances the weight of the electron is,

$E = - 5.57 \times {10^{ - 11}}\,{\rm{N/C}}$ .

As the sign of the electric field is negative, the direction of the electric field that balances the weight of the electron is downwards.

Ans: Part A

The strength of the electric field that will balance the weight of the proton is $1.023 \times {10^{ - 7}}\,{\rm{N/C}}$ .