Question

Oppositely charged parallel plates are separated by 5.31 mm. A potential difference of 600 V exists between the plates.
(a) What is the magnitude of the electric field between the plates?
N/C

(b) What is the magnitude of the force on an electron between the plates?
N

(c) How much work must be done on the electron to move it to the negative plate if it is initially positioned 2.95 mm from the positive plate?
J

Answer 1

Concepts and reason

The concepts needed to solve this problem are capacitance of a parallel plate capacitor, electric field, electric force and work done.

The relation between electric field, plate separation and potential difference is used to find the magnitude of the electric field between the plates.

Using the equation between force, electric field and charge of the electron, find the magnitude of the force on an electron between the plates.

The relation between work done, force and displacement can be used to find the work done on the electron to move it to the negative plate.

Fundamentals

When the plates of the parallel plate capacitor are connected to a power supply, one of the plates acquires a positive charge and the other plate becomes negatively charged. As the plates are oppositely charged, there is a net flow of charges from one plate to another, which is the electric field.

Thus, the source of the electric field is the potential difference. Electric field is the field produced by the charges and within this field any charged particle will experience a force. Electric field by definition is force per unit charge.

For a parallel plate capacitor, the electric field between the plates is,

$E = \frac{V}{d}$

Here, $V$ is the potential difference and $d$ is the separation between the plates.

The flow of electric field is from positive plate to the negative plate. Thus, a charge kept between the plates in the electric field experiences force called electric force. The electric force is the product of charge and the electric field.

The force on an electron between the plates is,

$F = qE$

Here, $q$ is the charge of an electron which is $1.6 \times {10^{ - 19}}\,{\rm{C}}$ and $E$ is the electric field between the plates.

When a force is applied to move a body, work is done. The work done to move a charge is the product of the force applied and the distance travelled by the charge.

The work done to move an electron from a position to another position in between the plates is,

$W = Fs$

Here, $F$ is the force and $s$ is the displacement.

(a)

The expression for the electric field is,

$E = \frac{V}{d}$

Substitute $600\,{\rm{V}}$ for $V$ and $5.31\,{\rm{mm}}$ for $d$ to find $E$ .

$\begin{array}{c}\\E = \frac{{600\,{\rm{V}}}}{{\left( {5.31\,{\rm{mm}}} \right)\left( {\frac{{{{10}^{ - 3}}\,{\rm{m}}}}{{1\,{\rm{mm}}}}} \right)}}\\\\ = 1.13 \times {10^{ - 6}}\,{\rm{N/C}}\\\end{array}$

(b)

The electric force on the electron is,

$F = qE$

Substitute $1.6 \times {10^{ - 19}}\,{\rm{C}}$ for $q$ and $1.13 \times {10^5}\,{\rm{N/C}}$ for $E$ to find $F$ .

$\begin{array}{c}\\F = \left( {1.6 \times {{10}^{ - 19}}\,{\rm{C}}} \right)\left( {1.13 \times {{10}^5}\,{\rm{N/C}}} \right)\\\\ = 1.808 \times {10^{ - 14}}\,{\rm{N}}\\\end{array}$

(c)

The work done is,

$W = Fs$

Displacement of the electron is,

${\rm{Displacement}} = {\rm{Final position}} - {\rm{Initial position}}$

Substitute $5.31\,{\rm{mm}}$ for final position and $2.95\,{\rm{mm}}$ for initial position to find displacement.

$\begin{array}{c}\\s = \;\left( {\left( {5.31\,{\rm{mm}}} \right)\left( {\frac{{{{10}^{ - 3}}\,{\rm{m}}}}{{1\,{\rm{mm}}}}} \right)} \right) - \left( {\left( {2.95\,{\rm{mm}}} \right)\left( {\frac{{{{10}^{ - 3}}\,{\rm{m}}}}{{1\,{\rm{mm}}}}} \right)} \right)\\\\ = 2.36 \times {10^{ - 3}}\,{\rm{m}}\\\end{array}$

Substitute $1.808 \times {10^{ - 14}}\,{\rm{N}}$ for $F$ and $2.36 \times {10^{ - 3}}\,{\rm{m}}$ for $s$ to find $W$ .

$\begin{array}{c}\\W = \left( {1.808 \times {{10}^{ - 14}}\,{\rm{N}}} \right)\left( {2.36 \times {{10}^{ - 3}}\,{\rm{m}}} \right)\\\\ = 4.27 \times {10^{ - 17}}\,{\rm{J}}\\\end{array}$

Ans: Part a

The magnitude of the electric field between the plates is $1.13 \times {10^5}\,{\rm{N/C}}$ .

Part b

The force on an electron between the plates is $1.808 \times {10^{ - 14}}\,{\rm{N}}$ .

Part c

The work done on the electron to move it to the negative plate if it is initially positioned $2.95\,{\rm{mm}}$ from the positive plate is $4.27 \times {10^{ - 17}}\,{\rm{J}}$ .