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18.3Electric potential and electric potential energy

Most of what we learned about gravitational potential energy and gravitational potential applies also to electricity. Just as a mass creates a gravitational potential around it, an electric charge creates electric potential the work done by an external agent per unit charge in bringing a small point positive charge from infinity to a point in an electric field. And just as two masses have gravitational potential energy between them, two electric charges also share electric potential energy the work that needs to be done by an external agent in order to bring a set of charges from where they were separated by an infinite distance to their current position. The formulas we derived for gravitation carry over to electricity essentially by replacing mass everywhere by charge, as we will see. The ideas are the same as those for gravitation so the derivations for electricity will be brief.

Suppose that at some point in space we place a large positive charge Q. If we place another positive charge q at infinity and try to move it closer to the large charge Q, we will have to exert a force on q, since it is being repelled by Q (Figure 18.64). That is, we have to do work in order to change the position of q and bring it closer to Q.

Figure 18.64: Work is done to bring the positive charge q from infinity to a given position away from positive charge Q. The red arrow is the force of repulsion between the two charges. The green arrow represents the force that moves charge q towards Q.

KEY POINT

The electric potential at a point P in an electric field is the amount of work done per unit charge as a small positive test charge q is moved from infinity to P

V e=W q

The unit of potential is the volt (V), and 1 V = 1 J C−1.

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The work done in moving a charge q from infinity to point P is the electric potential energy, E P of the system.

If the electric potential at some point P is V e, and we place a charge q at P, the electric potential energy E P of the system is given by:

E p=q V c

Using calculus to calculate the work done in moving the charge q from infinity to a separation r, as we did for gravitation, results in:

W=k Q q r

Therefore:

V e=k Q r

and

E p=k Q q r

Electric potential and electric potential energy are scalar quantities, just like gravitational potential and gravitational potential energy. For gravitational and electric fields, the work done is independent of the path followed.

Moving a charge q from one point in an electric field to another requires work (Figure 18.65).

Figure 18.65: Work must be done in order to move a charge from one point to another where the potential is different.

The work done W in moving charge q from A to B is:

W=q Δ V e=q(V eB−V e)

The quantity V eB – V eA is the potential difference between A and B.

In all these formulas, the charges must be entered with their correct sign.

Worked Example 18.10

The hydrogen atom has a single proton and a single electron, as shown in Figure 18.66.

Figure 18.66: For worked example 18.10.

aFind the electric potential a distance of 0.50 × 10−10 m from the proton of the hydrogen atom. The proton has a charge 1.6 × 10−19 C, equal and opposite to that of the electron.

bUse your answer to a to calculate the electric potential energy between the proton in a hydrogen atom and an electron orbiting the proton at a radius 0.50 × 10−10 m.

Answer

a V e=k Q r

Substituting the values from the question:

V e=8.99×10 9×1.60×10−19 0.50×10−10

V e=28.77≈29 V

bThe electric potential energy is given by:

E p=k Q q r=q V e

Substituting the value for V e from part a:

E p=28.77×(−1.6×10−19)

E p=−4.6×10−18 J

Electric potential is a scalar quantity. So if we have two charges q 1 and q 2, the electric potential at a point P that is a distance r 1 from q 1 and a distance r 2 from q 2 is just the sum of the individual electric potentials:

V e=k q 1 r 1+k q 2 r 2

That is, we first find the potential at P from q 1 alone, then from q 2 alone, and then add up the two (Figure 18.67). We find the electric potential for more than two charges in the same way—by adding the individual potentials.

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Figure 18.67: The potential at P is found by finding the potential there from the first charge, then finding the potential from the second charge, and finally adding the two.

The simple formula for electric potential works for point charges. (By point charges we mean that the objects on which the charges q i are placed are mathematical points, or close to it.) The formula also works in another special case—when the object on which the charge q is placed is a sphere.

For a point P outside the sphere and at a distance r from the centre of the sphere, the potential at P is indeed:

V e=k Q r

On the surface of the sphere the potential is:

V c=k Q R

where R is the radius of the sphere.

But at any point inside the sphere, the electric potential is constant and has the same value as the potential at the surface (Figure 18.68).

Figure 18.68: The electric potential is constant inside the sphere and falls off as 1 r outside. Shown here are a a positively charged sphere and b a negatively charged sphere.

The connection between potential and field

There is a deep connection between potential and field for both gravitation and electricity. To see this connection let us move a point charge q from a point P to a neighbouring point Q. There is a potential difference ∆V e between the two points, and the points are a distance ∆r apart, Figure 18.69.

Figure 18.69: A point charge q is to be moved between points P and Q at different potential. This requires work.

We know that this requires an amount of work W given by:

W=q Δ V e

But we may also calculate the work from W = force × distance. The force on the point mass is the electric force F = qE, where E is the magnitude of the electric field strength at the position of the charge q. Assuming that the two points are very close to each other means that E will not change very much as we move from one point to the other, and so we may take E to be constant. Then the work done is also given by:

W=q E Δ r

Equating the two expressions for work done gives:

E=Δ V°Δ r

(A more careful treatment using calculus gives E=−d V c d r.)

In a graph showing the variation with distance of the potential, the gradient (slope) of the graph is the magnitude of the field strength. This applies to both gravitational and electric fields.

KEY POINT

We conclude, see Figure 18.68, that the electric field inside a charged conducting sphere is zero since the potential is constant.

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Worked Example 18.11

Figure 18.70 shows two unequal positive charges, +Q and +q.

Figure 18.70: For worked example 18.11.

Which one of the four graphs in Figure 18.71 shows the variation with distance x from the larger charge of the electric potential V e along the line joining the centres of the charges?

Figure 18.71: Variation of electric potential with distance.

Answer

The electric field is the gradient of the graph. It is zero closer to the smaller charge, and so the answer has to be B.

THEORY OF KNOWLEDGE

Are fields real?

Are electric (and gravitational) fields real, or are they just convenient devices for doing calculations and visualizing situations?

Feynman describes a situation of two protons moving at right angles to each other:

The protons repel each other with the electric force (red arrows). These forces are equal in magnitude and opposite in direction. Proton 2 creates a magnetic field at the position of proton 1 directed into the plane of the page and so there is a magnetic force on proton 1 (green arrow). However, proton 1 does not create a magnetic field at the position of proton 2 because (at this instant) proton 2 is along the direction of the velocity of proton 1. Hence there is no magnetic force on proton 2. This appears to violate Newton’s third law but as we know Newton’s third law is crucial in deriving the law of conservation of momentum. It can be shown that there is a rate of change of the momentum carried by the electromagnetic field of the protons of the right magnitude and direction so as to restore Newton’s third law. Fields carry energy and momentum and are as real as protons and electrons.

The connection between field lines and equipotential surfaces

Consider a single charge Q. The electric potential it creates in the space around is given by V e=k Q r. This means that points in space that have the same distance from the charge have the same potential. These points lie on spheres centred at the charge. Spheres of different radii correspond to different values of the potential. We call surfaces of the same potential equipotential surfaces surfaces where all points are at the same potential. Figure 18.72a shows equipotential surfaces for a point charge (the surfaces are spheres surrounding the charge; we only show a two-dimensional version of them as circles here). Figure 18.72b adds field lines (the charge is assumed negative) along with the equipotential surfaces. We see that the field lines are normal (perpendicular) to the equipotential surfaces.

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Figure 18.72: a Equipotential surfaces for a point charge. b The field lines are at right angles to the equipotential surfaces.

The explanation of why the field lines are at right angles to the equipotential surfaces is as follows: to move a charge from one point on an equipotential surface to another requires zero work because W = qΔV and ΔV = 0. If the field lines were not normal to the equipotential surfaces, there would be a component of the field along the equipotential and so a force on the charge. As the charge moved this force would do work, which contradicts the fact that the work should be zero.

Equipotential surfaces are spheres only for an isolated charge. With more than one charge present, the equipotential surfaces are complicated surfaces. Figure 18.73 shows the equipotential surfaces (in black) and field lines (in red) for two charges. In Figure 18.73a q 2 = q 1 (both positive) and in Figure 18.73b q 2 = −q 1. The field is zero halfway between the charges along the line joining the charges in Figure 18.73a.

Figure 18.73: Field lines and equipotential surfaces for a two equal charges of the same sign and b two equal and opposite charges.

Figure 18.74 shows the equipotential surfaces (in black) and field lines (in red) for two unequal charges. The field is zero closer to the smaller charge along the line joining the charges in Figure 18.74a. In Figure 18.74a q 2 = 4q 1 and in Figure 18.74b q 2 = −4q 1.

Figure 18.74: Field lines and equipotential surfaces for a two unequal charges of the same sign and b two unequal and opposite charges.

Figure 18.75 shows equipotential surfaces and field lines for the case of parallel charged plates.

Figure 18.75: Field lines and equipotential surfaces for parallel charged plates.

Worked Example 18.12

A wire of length L has a potential difference V across its ends.

aFind the electric field inside the wire.

bHence find the work done when a charge q is moved from one end of the wire to the other.

Answer

aFrom E=Δ V Δ r it follows that E=V L.

bThe work done can be found in two ways.

1Use W=q Δ V=q V.

2Use W=F d=q E L=q V L L=q V.

The answer is W = qV in both cases.

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TEST YOUR UNDERSTANDING

38 aDetermine the electric potential at the mid-point of the line joining two equal positive charges q in terms of q, the Coulomb constant and the charge separation d.

bRepeat a for two equal but opposite charges.

39Two charges, q 1 = 2.0 µC and q 2 = −4.0 µC, are 0.30 m apart. Find the electric potential at a point P, which is 0.40 m from q 1 and 0.60 m from q 2.

40Four equal charges of 5.0 µC are placed at the vertices of a square of side 10 cm.

aCalculate the value of the electric potential at the centre of the square.

bDetermine the electric field at the centre of the square.

cHow do you reconcile your answers to a and b with the fact that the electric field is the derivative of the potential?

41A charge q of 10.0 C is placed somewhere in space.

aWhat is the work required to bring a charge of 1.0 mC from a point X, 10.0 m from q, to a point Y, 2.0 m from q?

bDoes the answer depend on which path the charge follows?

42An electron is brought from infinity to a distance of 12 cm from a charge of −15 nC. How much work was done on the electron by the agent moving the electron?

43An electron moves from point A where the potential is 100.0 V to point B where the potential is 200.0 V. The electron started from rest. Calculate the speed of the electron as it passes point B.

44A proton is placed on the surface of a positively charged sphere of charge 2.0 nC and radius 0.25 m. The proton is released.

aWhat is the speed of the proton when it gets very far away from the sphere?

bDraw a graph to show the variation with distance travelled, of the proton speed. (No numbers required, just the shape.)

45Four charges are placed at the vertices of a square of side 5.00 cm, as shown in Figure 18.76.

Figure 18.76: For question 45.

aOn a copy of Figure 18.76, show the forces acting on the 2.0 µC charge. Find the magnitude and direction of the net force on the 2.0 µC charge.

bCalculate the value of the electric potential at the centre of the square.

cDetermine the work that must be done in order to move a charge of 1 nC initially at infinity to the centre of the square.

46An electron is launched with speed 1.5 × 106 m s−1 from the positively charged plate, directed towards the opposite plate (Figure 18.77).

The potential difference between the plates is 8.0 V. The plates are separated by 12 cm. (The potential at the negative plate may be taken to be zero and that at the positive plate 8.0 V.)

Figure 18.77: For question 46.

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aShow that the electron will not make it to the negative plate.

bDetermine the distance at which the electron stops and turns back.

cWhat should the minimum electron initial speed be such that the electron makes it the negative plate?

dThe electron has the minimum speed found in c. Draw a sketch graph to show the variation of the electron speed with distance travelled.

47Two conducting spheres are separated by a distance that is large compared with their radii. The first sphere has a radius of 10.0 cm and has a charge of 2.00 µC on its surface. The second sphere has a radius of 15.0 cm and is neutral. The spheres are then connected by a long conducting wire.

aFind the charge on each sphere.

bCalculate the charge density on each sphere (charge density is the total charge on the sphere divided by the surface area of the sphere).

cCalculate the electric field on the surface of each sphere.

dComment on your result in the light of your answer to part b. Why is it stated that the wire is long?

48Figure 18.78 shows the equipotential lines for two equal and opposite charges. Draw the electric field lines for these two charges.

Figure 18.78: For question 48.

49Two long parallel plates are separated by a distance of 15.0 cm. The bottom plate is kept at a potential of −250 V and the top at +250 V. A charge of −2.00 µC is placed at a point 3.00 cm from the bottom plate.

aFind the electric potential energy of the charge.

The charge is then moved vertically up to a point 3.00 cm from the top plate.

bWhat is the electrical potential energy of the charge now?

cHow much work was done on the charge?

50An electron is shot with a speed equal to 1.59 × 106 m s−1 from a point where the electric potential is zero toward an immovable negative charge q (see Figure 18.79).

Figure 18.79: For question 50.

aDetermine the potential at P so that the electron stops momentarily at P and then turns back.

bCalculate the magnitude of q.

51Two equal and opposite charges are placed at points with coordinates x = 0, y = a and x = 0, y = −a, as shown in Figure 18.80.

Figure 18.80: For question 51.

aFind the electric field at the point with coordinates x = d, y = 0.

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bRepeat for two equal negative charges −q on the y-axis.

cSketch graphs to show the variation of these fields with the distance d.

52Three protons are initially very far apart. Calculate the work that must be done in order to bring these protons to the vertices of an equilateral triangle of side 5.0 × 10−15 m.

53A charge −q whose mass is m moves in a circle of radius r around another positive stationary charge q located at the centre of the circle, as shown in Figure 18.81.

Figure 18.81: For question 53.

aDraw the force on the moving charge.

bShow that the velocity of the charge is given by:

v 2=k q 2 m r

cShow that the total energy of the charge is given by:

E=−k q 2 2 r

dHence determine how much energy must be supplied to the charge if it is to orbit around the stationary charge at a radius equal to 2r.

54An electron of charge −e and mass m orbits the proton in a hydrogen atom as in the previous problem.

aShow that the period of revolution of the electron is given by T 2=4 π 2 m k e 2 r 3 where k is the Coulomb constant and r the radius of the orbit.

bCalculate this period for an orbit radius of 0.5 × 10−10 m.

cUsing the results of the previous problem calculate the energy that must be supplied to the electron so it orbits the proton in an orbit of radius 2.0 × 10−10 m.

55An alpha particle (mass = 6.6 × 10−27 kg, charge = +2e) is directed towards a nucleus of electric charge +79e. The radius of this nucleus is 7.0 × 10−15 m. Initially the alpha particle is very far from the nucleus. What should the initial speed of the alpha particle be so that it just stops on the nuclear surface?