electricity

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Electrostatic Potential Energy

To put two positive charges \(q_1\) and \(q_2\) together, we have to do work. This is called electrostatic potential energy.

To place \(q_1\) in an empty place doesn’t take any work. But to then bring \(q_2\) from \(\infty\) to point \(P\) at distance \(R\), we have to push with force \(F_{wl}\) to overcome the force of the electric field \(F_{el}\) from \(q_1\) on \(q_2\).

The force and the direction we’re moving are the same, so we do positive work. The work done to bring \(q_2\) from \(\infty\) to \(P\), is the same as minus the electric force $$ \shaded{ W_{wl} = \int_\infty^R \vec F_{wl}\cdot d\vec r = \int_R^\infty \vec F_{el}\cdot d\vec r } \tag{Work} $$

The electric force follows from Coulomb’s law

$$ F_{el} = \frac{q_1\,q_2\,K}{r^2} \nonumber $$

\(\newcommand{\parallelsum}{\mathbin{\!/\mkern-5mu/\!}} \vec F_{el} \parallelsum d\vec r \), so the dot-product becomes a multiplication of scalars \(|\vec F_{el}|\,|d\vec r|\,\cos 0=F_{el}\,dr\) $$ \begin{align*} W_{wl} &= \int_R^\infty \frac{q_1q_2}{4\pi\varepsilon_0r^2}\,dr = \frac{q_1q_2}{4\pi\varepsilon_0} \int_R^\infty \frac{dr}{r^2} \\ &= \frac{q_1q_2}{4\pi\varepsilon_0} \left[-\frac{1}{r}\right]_R^\infty \end{align*} \nonumber $$

The energy \(U_P\) (\(E_{pot}\) in Europe) required to bring \(q_2\) from \(\infty\) to point \(P\) $$ \shaded{ U_P = \frac{q_1q_2}{4\pi\varepsilon_0R} } \quad \left[\rm{J}\right] \tag{Potential energy} $$

Notes

• If we came over a very crooked line, the amount of work would be the same (for a conservative force, the work is independent of the path).
• With multiple charges, the superposition principle applies.

Work is the difference between the potential energy at the start point and the end point. $$ \shaded{ W = U_1 – U_2 } \tag{Work} $$

Electrostatic Potential

Assume one charge \(Q\), a test charge \(q\) at distance \(\infty\), and point \(P\) at distance \(R\) from \(Q\).

We just calculated the electrostatic potential energy \(U_P\), the work required to bring \(q\) from \(\infty\) to point \(P\)

$$ U_P = \frac{qQ}{4\pi\varepsilon_0 R} \nonumber $$

The electric potential \(V_P\) (\(\phi\) in Europe) is defined as the work per unit charge required to bring test charge \(q\) from \(\infty\) to point \(P\) $$ \shaded{ V_P = \frac{U_P}{q} = \frac{Q}{4\pi\varepsilon_0 R} } \quad \left[\rm{V=\frac{J}{C}}\right] \tag{Potential} $$

Note

• The unit is Volt named after Alessandro Giuseppe Antonio Anastasio Volta, an Italian physicist, chemist, who invented the battery.
• If \(Q\) is negative, then \(V_P\) is negative.
• \(V_P\) is the potential at distance \(R\) from charge \(Q\).
• \(V_\infty = 0\)

Example: van der Graaff

A van de Graaff generator with radius \(R\) and charge \(Q\). What is the electric potential at any point in space?

We already know

• inside the sphere, \(E=0\),
• the charge behaves as a point charge in the center of the sphere.

The potential at point \(P\) is defined as $$ \shaded{ V_P = \int_P^\infty \frac{\vec F_{el}}{q}\cdot d\vec r = \int_P^\infty \vec E\cdot d\vec r } \tag{Potential} \label{eq:electpotential} $$

Both \(\vec E\) and \(\vec r\) are radial, so the dot-product becomes a multiplication of scalars \(|\vec E|\,|d\vec r|\,\cos 0=E.dr.1\); substitute \(\vec E\) from what we found earlier $$ \begin{align*} V_P &= \int_R^\infty \frac{Q}{4\pi r^2\varepsilon_0} dr = \frac{Q}{4\pi\varepsilon_0} \int_R^\infty \frac{1}{r^2}\,dr \\ &= \frac{Q}{4\pi\varepsilon_0} \left[-\frac{1}{r}\right]_R^\infty = \frac{Q}{4\pi\varepsilon_0 r} & (r\gt R) \end{align*} \nonumber $$

Note

• This is of course the same as for the point charge.
• For \(r=R\), \(V=300\,\rm{kV}\); for \(r=3m\), \(V=30\,\rm{kV}\).
• If I bring charge \(q\) from \(\infty\) to the surface of the van der Graaff, I do work \(W_R = q\,V_R\)

If bring the charge inside the van der Graaff, where the field is \(0\), I don’t experience any force \(\Longrightarrow\) I do no work \(\Longrightarrow\) the potential must remain constant, \(V_R\). The absence of an electric field implies that the potential everywhere is the same.

Equipotential surfaces

Two charges \(Q_1\) and \(Q_2\). What is the potential at point \(P\)?

The superposition principle applies $$ V_P = V_{P_{Q1\text{ alone}}} + V_{P_{Q2\text{ alone}}} \nonumber $$

Example: two positive charges

An example of equipotential lines (and field lines)

Note

• If you go far away and close to the charges, the equipotential lines become spheres.
• In between the charges, there is a point where the electric field is \(0\), but has a positive potential.
• Everywhere the field lines are perpendicular to the equipotential lines. Imagine you purposely move with a charge only perpendicular to the electric field lines \(\Longrightarrow\) the force on you and the direction in which you move are always at a 90° angle \(\Longrightarrow\) you never do any work, because \(\vec E\cdot d\vec l=0\) \(\Longrightarrow\) the potential remains the same.

  

  These is really 3D, so you should rotate it around the vertical

• The lines are plotted as quantiles, to make it less cluttered around the charges.

One positive and one negative charge

Another example of equipotential lines (and field lines)

Note

• At one of the lines around \(-1\) the potential has to be \(0\), because inside it is negative, and far away it is positive. If you come from far away, and arrive at that \(0\) line, you have done \(0\) work. \(E\neq 0\) there
• Above the \(-1\), there is a point where the electric field is \(0\). \(V\neq 0\) there

In cases where the electric field is very complicated, it is easier to work with the potentials. The change in kinetic energy, only depends on the change in potential.

If you’re only interested in the change of kinetic energy, and not interested in the details of the trajectory, then equipotential lines are very handy.

Do not confuse:

• electrostatic potential energy \(U\) with unit Joules,
• with electric potential \(V\) with unit Volts (Joules/Coulomb).

In gravity, matter want to go from high potential to a place with low potential. Analogue, positive charges will also go from a high electric potential to a low electric potential. Uniquely for electricity, negative charges will go from a low potential to a high potential.

Potential difference

Two points in space, with a potential, separated a distance \(R\)

The potentials \(V_A\) and \(V_B\) are defined in equation \(\eqref{eq:electpotential}\) as $$ \begin{align*} V_A &= \int_A^\infty \vec E\cdot d\vec r \\ V_B &= \int_B^\infty \vec E\cdot d\vec r \end{align*} $$

\(\vec E\) is force per unit charge, not work.

So, the potential difference between point \(A\) and \(B\) $$ \begin{align*} V_A – V_B &= \int_A^B \vec E\cdot d\vec r \\ V_B – V_A &= -\int_A^B \vec E\cdot d\vec r \end{align*} \nonumber $$

Note

• If there is no electric field between \(A\) and \(B\), the potential is the same.

We change the \(d\vec r\) to a different symbol: element \(d\vec l\) to indicate any path between \(A\) and \(B\)

It makes no difference how you march, because we’re dealing with conservative fields (independent of the path).

The potential difference \(\Delta V\) (\(U\) in Europe) $$ \shaded{ \Delta V \overset{\Delta}{=} V_A – V_B = \int_A^B \vec E\cdot d\vec l } \tag{Potential difference} \label{eq:electpotdiff} $$

Some books write this as $$ V_B – V_A = -\int_A^B \vec E\cdot d\vec l \nonumber $$

\(U\)Instead of \(\Delta V\), we often use the European symbol for voltage difference \(U\). The letter ‘U’ stands for “Potentialunterschied”.

Let’s assume \(V_A=150\,\rm V\) and \(V_B=50\,\rm V\)

• If I bring a positive charge \(q\) to point \(A\), I have to do work \(q\,V_B = 50\,q\)
• If I bring the charge \(q\) to point \(B\), I have to do work \(q\,V_A = 150\,q\)

If I am at point \(A\) the charge \(q\) go to the lower potential at point \(B\), and \(150\,q-50\,q\) kinetic energy is released. The change in potential energy, when the plus \(q\) charge goes from \(A\) to \(B\), is \(V_A – V_B\) $$ q(V_A-V_B) \nonumber $$

In conservative fields, the sum of potential and kinetic energy stays the same. The conservation of energy. So $$ q(V_A-V_B) = K_B – K_A \nonumber $$ where \(K\) is kinetic energy.

Conductors

Any piece of metal is an equipotential, as long as there no charge moving inside the metal. Because these charges inside the metal, these electrons, when they experience an electric field, they move immediately in the electric field. They will move until there is no force on them any more \(\Longrightarrow\) they have made the electric field \(0\).

Charges inside a conductor always move automatically in such a way that they kill the electric field inside.

Metals, for as long as there is no charge moving inside, will always be an equipotential.

Example 1

If I connect point \(A\) to a metal trash can, and point \(B\) to a metal soda can, then the trash can will be at \(150\,\rm V\) and the whole soda can will be at \(50\,\rm V\). I place the whole thing in vacuum, and release an electron at point \(B\).

The electron wants to go to a higher potential. The electric potential energy is available, and the electron will start to pick up speed and end up at \(A\). How it will travel, I don’t know. The field configuration is enormously complicated between the soda can and the trash can.

But if all we want to know is the kinetic energy, the speed, then we can use the work-energy theorem. The available potential energy is charge of the electron time the potential difference between two objects $$ \begin{align*} q (V_A-V_B) &= K_B – K_A \\ \Rightarrow -1.6\times 10^{-19} (100) &= 0 – \tfrac{1}{2}m_e {v_A}^2 \\ \Rightarrow 1.6\times 10^{-19} (100) &= \tfrac{1}{2}9\times10^{-31}\,v{_A}^2 \\ \Rightarrow v_A &\approx 5.96\times 10^6\,\rm\frac{m}{s}\approx 0.02\,c \end{align*} $$

Using the equipotential, we can quickly calculate the kinetic energy and the speed of the electron as it arrives at \(A\), without any knowledge of the complicated electric field.

All potentials are defined relative to \(\infty\) \(\Longrightarrow\) \(V_\infty=0\). But it doesn’t matter where you think of you \(0\), similar as the height with gravity. Only the potential difference matters. Note that electrical engineers always call the potential of the earth \(0\) when they build their circuits.

Example 2 (Lec.12)

Kinetic energy increase due to charges that move over a potential difference.

Two conductors of odd shape that are equipotential (no current inside the conductors) in vacuum. Let’s assume \(V_A\gt V_B\)

Release charge \(q\) at speed zero, that will go to \(B\). What is the speed as it reaches \(B\)?

The electric field will do the work \(W_{el}\) on this charge. Since it is a conservative field, it is path independent. $$ \begin{align*} W_{el} &= \int_A^B \vec F_{el}\cdot d\vec l, & \left(\vec F = q\vec E \right) \\ &= q \int_A^B \vec E\cdot d\vec l \\ \end{align*} $$

Where \(\int_A^B\vec E\cdot d\vec l\) is the potential difference $$ W_{el} = q\,(V_A – V_B) \nonumber $$

Quantitively

A proton traverses a potential difference $$ \begin{align*} m_p &= 1.7\times 10^{-27}\,\rm{kg} \\ q_p &= 1.6\times 10^{-19}\,\rm{C} \\ \Delta V &= 10^6\,\rm V \end{align*} $$

What is the kinetic energy when the proton reaches \(B\)? $$ \begin{align*} W_{el} &= q\,\Delta V \\ &= 1.6\times 10^{-19}\,(10^6) \\ &= 1.6\times 10^{-13}\,\rm J \end{align*} $$

This is commonly called one million electro volt, \(1\,\rm{MeV}\). Where an electro volt is the energy that an electron gains if it moves over a potential difference of one volt.

The speed that it arrives at \(B\) $$ \begin{align*} W_{el }&= \tfrac{1}{2} m_p\,v^2 \\ \Rightarrow v &= \sqrt\frac{2W_{el}}{m_p} \\ &= \sqrt\frac{2\left(1.6\times 10^{-13}\right)}{1.7\times 10^{-27}} \\ &\approx 1.37\times 10^7\,\rm{m/s} \approx 0.05\,c \end{align*} $$

One more, that is 5% of the speed of light, comfortably low, so we don’t have to make any relativistic corrections.