FHSST Physics/Electrostatics/Electrical Potential

Electrical Potential

Work Done and Energy Transfer in a Field

When a charged particle moves in an electric field work is done and energy transfers take place. This is exactly analogous to the case when a mass moves in a gravitational field such as that set up by any massive object.

Work done by a field

Gravitational Case

A mass held at a height h above the ground has gravitational potential energy since, if released, it will fall under the action of the gravitational field. Once released, in the absence of friction, only the force of gravity acts on the mass and the mass accelerates in the direction of the force (towards the Earth's centre).

In this way, work is done by the field. When the mass falls a distance h (from point A to B), the work done is,

${\begin{matrix}W&=&Fs\\&=&mgh\end{matrix}}$

In falling, the mass loses gravitational potential energy and gains kinetic energy.

Energy is conserved!

The work done by the field is equal to the energy transferred,

${\begin{matrix}W=\mathrm {Gain\ in\ } E_{k}=\mathrm {Loss\ in\ } E_{p}\qquad \mathrm {(a\ falling\ mass)} \end{matrix}}$

Electrical Case

A charge in an electric field has electrical potential energy since, if released, it will move under the action of the electric field. When released, in the absence of friction, only the electric force acts on the charge and the charge accelerates in the direction of the force (for positive charges the force and acceleration are in the direction of the electric field, while negative charges experience a force and acceleration in the opposite direction to the electric field.) Consider a positive charge +Q placed in the uniform electric field between oppositely charged parallel plates.

The positive charge will be repelled by the positive plate and attracted by the negative plate (i.e. it will move in the direction of the electric field lines). In this way, work is done by the field. In moving the charge a distance s in the electric field, the work done is,

${\begin{matrix}W&=&Fs\\&=&QEs\qquad \mathrm {since\ } E={F \over Q}.\end{matrix}}$

In the process of moving, the charge loses electrical potential energy and gains kinetic energy. The work done by the field is equal to the energy transferred,

${\begin{matrix}W&=&\mathrm {Gain\ in\ } E_{k}=\mathrm {Loss\ in\ } E_{p}\qquad \mathrm {(charge\ moving\ under\ the\ influence\ of\ an\ electric\ field)} \end{matrix}}$

Work done by us

Gravitational Case

In order to return the mass m in Figure 12.1 to its original position (i.e. lift it a distance h from B back to A) we have to apply a force mg to balance the force of gravity. An amount of work mgh is done by the lifter. In the process, the mass gains gravitational potential energy,

${\begin{matrix}mgh&=&\mathrm {Gain\ in\ } E_{p}\qquad \mathrm {(lifting\ a\ mass)} \end{matrix}}$

Electrical Case

In order to return the charge in Figure 12.2 to its original position (i.e. from B back to A) we have to exert a force QE on the charge to balance the force exerted on it by the electric field. An amount of work QEs is done by us. In the process, the charge gains electrical potential energy,

${\begin{matrix}QEs&=&\mathrm {Gain\ in\ } E_{p}\qquad \mathrm {(charge\ moved\ against\ an\ electric\ field)} \end{matrix}}$

In summary, when an object moves under the influence of a field, the field does work and potential energy is transferred into kinetic energy. Potential energy is lost, while kinetic energy is gained. When an object is moved against a field we have to do work and the object gains potential energy.

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Worked Example 62 Work done and energy transfers in a field

Question: A charge of +5nC is moved a distance of 4 cm against a uniform electric field of magnitude $2\times 10^{12}\mathrm {N.C^{-1}}$ from A to B.

(a) Calculate the work done in moving the charge from A to B. (b) The charge is now released and returns to A. Calculate the kinetic energy of the charge at A.

Answer: (a) Step 1 :

We are given the values of the charge, the field and the distance the charge must move. All are in the correct units.

Step 2 :

Since the charge is positive we have to do work to move it from A to B (since this is against the field). This work is given by,

${\begin{matrix}W&=&QEs\\&=&(5\times 10^{-9})(2\times 10^{12})(0.04)\\&=&400\ \mathrm {J} \end{matrix}}$

(b)

Step 3 :

When released the charge moves under the influence of the electric field and returns to A. Work is now done by the field and the work done is equal to the kinetic energy gained,

${\begin{matrix}\mathrm {Gain\ in\ } E_{k}&=&QEs\\&=&(5\times 10^{-9})(2\times 10^{12})(0.04)\\&=&400\ \mathrm {J} \end{matrix}}$

Since the charge started at rest, the gain in kinetic energy is the final kinetic energy,

${\begin{matrix}E_{k}^{\mathrm {at\ A} }&=&400\ \mathrm {J} \end{matrix}}$

Electrical Potential Difference

Consider a positive test charge + Q placed at A in the electric field of another positive point charge.

The test charge moves towards B under the influence of the electric field of the other charge. In the process the test charge loses electrical potential energy and gains kinetic energy. Thus, at A, the test charge has more potential energy than at B - A is said to have a higher electrical potential than B. The potential energy of a charge at a point in a field is defined as the work required to move that charge from infinity to that point.

The potential difference between two points in an electric field is defined as the work required to move a unit positive test charge from the point of lower potential to that of higher potential. If an amount of work W is required to move a charge Q from one point to another, then the potential difference between the two points is given by,

${\begin{matrix}V&=&{W \over Q}\qquad \mathrm {unit:J.C^{-1}\ or\ V\ (the\ volt)} \end{matrix}}$

From this equation it follows that one volt is the potential difference between two points in an electric field if one joule of work is done in moving one coulomb of charge from the one point to the other.

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Worked Example 63 Potential difference

Question: A positively charged object Q is placed as shown in the sketch. The potential difference between two points A and B is $4\times 10^{-4}\ \mathrm {V}$

(a) Calculate the change in electrical potential energy of a +2nC charge when it moves from A to B. (b) Which point, A or B, is at the higher electrical potential? Explain. (c) If this charge were replaced with another of charge -2nC, in what way would its change in energy be affected?

Answer: (a) The electrical potential energy of the positive charge decreases as it moves from A to B since it is moving in the direction of the electric field produced by the object Q. This loss in potential energy is equal to the work done by the field,

${\begin{matrix}\mathrm {Loss\ in\ Electrical\ Potential\ Energy} &=&W\\&=&VQ\qquad \mathrm {(Since\ } V={W \over Q})\\&=&(4\times 10^{-4})(2\times 10^{-9})\\&=&8\times 10^{-13}\ \mathrm {J} \end{matrix}}$

(b) Point A is at the higher electrical potential since work is required by us to move a positive test charge from B to A. (c) If the charge is replaced by one of negative charge, the electrical potential energy of the charge will increase in moving from A to B (in this case we would have to do work on the charge).

As an example consider the electric field between two oppositely charged parallel plates a distance d apart maintained at a potential difference V.

This electric field is uniform so that a charge placed anywhere between the plates will experience the same force. Consider a positive test charge Q placed at point O just off the surface of the negative plate. In order to move it towards the positive plate we have to apply a force QE. The work done in moving the charge from the negative to the positive plate is,

${\begin{matrix}W&=&Fs\\&=&QEd,\end{matrix}}$

but from the definition of electrical potential,

${\begin{matrix}W&=&VQ.\end{matrix}}$

Equating these two expressions for the work done,

${\begin{matrix}QEd&=&VQ,\end{matrix}}$

and so, rearranging,

${\begin{matrix}E&=&{V \over d}.\qquad \end{matrix}}$

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Worked Example 64 Parallel plates

Question: Two charged parallel plates are at a distance of 180 mm from each other. The potential difference between them is $3600\ \mathrm {V}$ as shown in the diagram.

(a) If a small oil drop of negligible mass, carrying a charge of $+6.8\times 10^{-9}\ \mathrm {C}$, is placed between the plates at point X, calculate the magnitude and direction of the electrostatic force exerted on the droplet.

(b) If the droplet is now moved to point Y, would the force exerted on it be bigger, smaller or the same as in (a)?

Answer: (a)

Step 1 :

First find the electric field strength between the plates,

${\begin{matrix}E&=&{V \over d}\\&=&{3600 \over 0.180}\\&=&20000\ \mathrm {N.C^{-1}\ from\ the\ positive\ to\ the\ negative\ plate} \end{matrix}}$

Step 2 :

Now the force exerted on the charge at X is,

${\begin{matrix}F&=&QE\\&=&(6.8\times 10^{-9})(20000)\\&=&1.36\times 10^{-4}\ \mathrm {down} \end{matrix}}$

(b)

Figure 12.3: An oil drop suspended between oppositely charged parallel plates.

File:Fhsst electrost29.png

Step 3 :

The same. Since the electric field strength is uniform, the force exerted on a charge is the same at all points between the plates.

Millikan's Oil-drop Experiment

Robert Millikan measured the charge on an electron by studying the motion of charged oil drops between oppositely charged parallel plates.

Consider one such negative drop between the plates in Figure 12.3. Since this drop is negative, the electric field exerts an upward force on the drop. In addition to this upward force, gravity exerts a downward force on the drop. Millikan adjusted the electric field strength between the plates by varying the potential difference applied across the plates. In this way, Millikan was able to bring the drops to rest. *EDIT* Millikan's experiment was meant to have the drops fall at a constant rate. At this constant rate the force of the earth on the drop and the force of the field on the drop are equal *End EDIT*

${\begin{matrix}F_{\mathrm {up} }&=&F_{\mathrm {down} }\\QE&=&mg\end{matrix}}$

Since $E={V \over d}$

${\begin{matrix}Q{V \over d}&=&mg,\\\end{matrix}}$

and, therefore,

${\begin{matrix}Q&=&{mgd \over V}\end{matrix}}$

Millikan found that all drops had charges which were multiples of $1.6\times 10^{-19}\ \mathrm {C}$. Since objects become charged by gaining or losing electrons, the charge on an electron must be $-1.6\times 10^{-19}\ \mathrm {C}$. The magnitude of the electron's charge is denoted by e,

${\begin{matrix}e&=&1.6\times 10^{-19}\ \mathrm {C} \end{matrix}}$

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Worked Example 65 Charge

Question: A metal sphere carries a charge of $+3.2\times 10^{-8}\ \mathrm {C}$. How many electrons did it have to lose to attain its charge?

Answer: Since the sphere is positive it lost electrons in the process of charging (when an object loses negative charges it is left positive). In fact, it lost,

${\begin{matrix}{3.2\times 10^{-8} \over 1.6\times 10^{-19}}&=&2\times 10^{11}\ \mathrm {electrons} \end{matrix}}$

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Worked Example 66 Millikan oil-drop experiment

Question: In a Millikan-type experiment a positively charged oil drop is placed between two horizontal plates, 20 mm apart, as shown.

The potential difference across the plates is 4000V. The drop has a mass of $1.2\times 10^{-14}$ and a charge of $8\times 10^{-19}$. (a) Draw the electric field pattern between the two plates. (b) Calculate:

1. the electric field intensity between the two plates.
2. the magnitude of the gravitational force acting on the drop.
3. the magnitude of the Coulomb force acting on the drop.

(c) The drop is observed through a microscope. What will the drop be seen to do? Explain. (d) Without any further calculations, give two methods that could be used to make the drop remain in a fixed position.

Answer: (a)

(b) 1.

${\begin{matrix}E&=&{V \over d}\\&=&{4000 \over 0.02}\\&=&2\times 10^{5}\ \mathrm {V.m^{-1}\ up} \\\end{matrix}}$

2.

${\begin{matrix}F_{\mathrm {grav} }&=&mg\\&=&(1.2\times 10^{-14})(10)\\&=&1.2\times 10^{-13}\mathrm {N} \end{matrix}}$

3.

${\begin{matrix}F_{\mathrm {Coulomb} }&=&QE\\&=&(8\times 10^{-19})(s\times 10^{5})\\&=&1.6\times 10^{-13}\mathrm {N} \end{matrix}}$

(c) Since $F_{up}>F_{down}$, the drop accelerates upwards.

(d) The Coulomb force can be decreased by decreasing the electric field strength between the plates. Since $E={V \over d}$, this can be done either by increasing d or decreasing V.