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# 12.1: Electricity

Difficulty Level: At Grade Created by: CK-12

## The Coulomb Force Law

The Coulomb Force Law states that any two charged particles \begin{align*} (q_1,q_2) \end{align*} --- with charge measured in units of Coulombs --- at a distance \begin{align*} r \end{align*} from each other will experience a force of repulsion or attraction along the line joining them equal to:

\begin{align*}\vec{F_e} = \frac{kq_1q_2}{r^2}&& \text{The Coulomb Force [1]}\\ \intertext{Where}\\ k = 8.987 \times 10^9 \ \mathrm{N \cdot m^2 \cdot C^{-2}}. &&\text{The Electric Constant}\end{align*}

This looks a lot like the Law of Universal Gravitation, which deals with attraction between objects with mass. The big difference is that while any two masses experience mutual attraction, two charges can either attract or repel each other, depending on whether the signs of their charges are alike:

Like gravitational (and all other) forces, Coulomb forces add as vectors. Thus to find the force on a charge from an arrangement of charges, one needs to find the vector sum of the force from each charge in the arrangement.

### Example 1

Question: Two negatively charged spheres (one with \begin{align*}-12\mu\mathrm{C}\end{align*}; the other with \begin{align*}-3\mu\mathrm{C}\end{align*}) are \begin{align*}3\mathrm{m}\end{align*} apart. Where could you place an electron so that it will be suspended in space between them with a net force of zero (for this problem we will ignore the force of repulsion between the two charges because they are held in place)?

Answer: Consider the diagram above; here \begin{align*} r_{s\rightarrow e} \end{align*} is the distance between the electron and the small charge, while \begin{align*}\vec{F}_{s\rightarrow e} \end{align*} is the force the electron feels due to it. For the electron to be balanced in between the two charges, the forces of repulsion caused by the two charges on the electron would have to be balanced. To do this, we will set the equation for the force exerted by two charges on each other equal and solve for a distance ratio. We will denote the difference between the charges through the subscripts “s” for the smaller charge, “\begin{align*}e\end{align*}” for the electron, and “\begin{align*}l\end{align*}” for the larger charge.

\begin{align*}\frac{kq_sq_e}{r_{s\rightarrow e}^2}=\frac{kq_lq_e}{r_{e\rightarrow l}^2}\end{align*}

Now we can cancel. The charge of the electron cancels. The constant \begin{align*}k\end{align*} also cancels. We can then replace the large and small charges with the numbers. This leaves us with the distances. We can then manipulate the equation to produce a ratio of the distances.

\begin{align*}\frac{-3\mu\mathrm{C}}{r_{s\rightarrow e}^2}=\frac{-3\mu\mathrm{C}}{r_{e\rightarrow l}^2} \Rightarrow \frac{r_{s\rightarrow e}^2}{r_{e\rightarrow l}^2}=\frac{-12\mu\mathrm{C}}{-12\mu\mathrm{C}} \Rightarrow \frac{r_{s\rightarrow e}}{r_{e\rightarrow l}}=\sqrt{\frac{1\mu\mathrm{C}}{4\mu\mathrm{C}}}=\frac{1}{2}\end{align*}

Given this ratio, we know that the electron is twice as far from the large charge (\begin{align*}-12\mu\mathrm{C}\end{align*}) as from the small charge (\begin{align*}-12\mu\mathrm{C}\end{align*}). Given that the distance between the small and large charges is \begin{align*}3\mathrm{m}\end{align*}, we can determine that the electron must be located \begin{align*}2\mathrm{m}\end{align*} away from the large charge and \begin{align*}1\mathrm{m}\end{align*} away from the smaller charge.

## Electric Fields and Electric Forces

Gravity and the Coulomb force have a nice property in common: they can be represented by fields. Fields are a kind of bookkeeping tool used to keep track of forces. Take the electromagnetic force between two charges given above:

\begin{align*}\vec{F_e} = \frac{kq_1q_2}{r^2}\end{align*}

If we are interested in the acceleration of the first charge only --- due to the force from the second charge --- we can rewrite this force as the product of \begin{align*}{q_1} \end{align*} and \begin{align*}\frac{kq_2}{r^2}\end{align*}. The first part of this product only depends on properties of the object we're interested in (the first charge), and the second part can be thought of as a property of the point in space where that object is.

In fact, the quantity \begin{align*}\frac{kq_2}{r^2}\end{align*} captures everything about the electromagnetic force on any object possible at a distance r from \begin{align*} q_2 \end{align*}. If we had replaced \begin{align*} q_1 \end{align*} with a different charge, \begin{align*} q_3 \end{align*}, we would simply multiply \begin{align*} q_3 \end{align*} by \begin{align*}\frac{kq_2}{r^2}\end{align*} to find the new force on the new charge. Such a quantity, \begin{align*}\frac{kq_2}{r^2}\end{align*} here, is referred to as the electric field from charge \begin{align*} q_2 \end{align*} at that point: in this case, it is the electric field due to a single charge:

\begin{align*}\vec{E_f} = \frac{kq}{r^2} &&\text{[2] Electric field due to point charge } q ,\text{ distance } r \text{ away}\end{align*}

The electric field is a vector quantity, and points in the direction that a force felt by a positive charge at that point would. If we are given the electric field at some point, it is just a matter of multiplication --- as illustrated above --- to find the force any charge \begin{align*} q_0 \end{align*} would feel at that point:

\begin{align*}\underbrace{\vec{F_e}}_{\text{Force on charge }q_0} = \underbrace{\vec{E_f}}_{\text{Field}} \underbrace{\times q_0}_{\text{Charge}} &&\text{Force on charge }q_0\text{ in an electric field}\end{align*}

Note that this is true for all electric fields, not just those from point charges. In general, the electric field at a point is the force a positive test charge of magnitude 1 would feel at that point. Any other charge will feel a force along the same line (but possibly in the other direction) in proportion to its magnitude. In other words, the electric field can be thought of as “force per unit charge”.

In the case given above, the field was due to a single charge. Such a field is shown in the figure below. Notice that this a field due to a positive charge, since the field arrows are pointing outward. The field produced by a point charge will be radially symmetric i.e., the strength of the field only depends on the distance, \begin{align*} r \end{align*}, from the charge, not the direction; the lengths of the arrows represent the strength of the field.

### Example 3

Question: Calculate the electric field a distance of \begin{align*}4.0\mathrm{mm}\end{align*} away from a \begin{align*}-2.0\mu \mathrm{C}\end{align*} charge. Then, calculate the force on a \begin{align*}-8.0\mu \mathrm{C}\end{align*} charge placed at this point.

Answer: To calculate the electric field we will use the equation

\begin{align*}E=\frac{kq}{r^2}\end{align*}

Before we solve for the electric field by plugging in the values, we convert all of the values to the same units.

\begin{align*}& 4.0\mathrm{mm}\times \frac{1\mathrm{m}}{1000\mathrm{mm}}=.004\mathrm{m}\\ & -2.0\mu \mathrm{C}\times \frac{1\mathrm{C}}{1000000\mu \mathrm{C}}=-2.0\times 10^{-6}\mathrm{C}\end{align*}

Now that we have consistent units we can solve the problem.

\begin{align*}E=\frac{kq}{r^2}=\frac{9\times 10^9\mathrm{Nm^2/C^2}\times -2.0\times 10^{-6}\mathrm{C}}{(.004\mathrm{m})^2}=-1.1\times 10^9\mathrm{N/C}\end{align*}

To solve for the force at the point we will use the equation

\begin{align*}F=Eq\end{align*}

We already know all of the values so all we have to do is convert all of the values to the same units and then plug in the values.

\begin{align*}& -8.0\mu \mathrm{C}\times \frac{1\mathrm{C}}{1000000\mu \mathrm{C}}=-8.0\times 10^{-6}\mathrm{C}\\ & F=Eq=-8.0\times 10^{-6}\mathrm{C} \times -1.1 \times 10^9 \mathrm{N/C}=9000\mathrm{N}\end{align*}

## Fields Due to Several Charges

To find the field at a point due to an arrangement of charges --- in fact, all electric fields arise due to some arrangement of charges --- we find the vector sum of the individual fields:

\begin{align*}\vec{E_{net}} = \sum_i {\vec{E_i}} &&\text{[3] Net Electric Field} \end{align*}

Electric fields are used more frequently than gravitational ones because there are two types of charge, which makes electric force and potential energy harder to keep track of than their gravitational counterparts. To apply this approach to gravitational forces --- that is, to find a net gravitational field --- one needs to repeat the steps above, with mass in place of charge (left for the reader).

### Example 4

Question: For the diagram above, draw (qualitatively) the electric field vectors at the points shown using the test charge method.

Answer: We will start with Test Charge 1. Test charges are always positive and have magnitude 1. Therefore we know that the test charge will want to go toward the negative charge and away from the positive charge (like charges repel and opposite charges attract). The strength of the electric field felt by the test charge is dependent on the inverse square of the distance of the charges as shown by the equation

\begin{align*}E=\frac{kq}{r^2}\end{align*}

The farther away from the source of the field, the weaker the field becomes. Therefore Test Charge 1 will experience a stronger field from the 1C charge. Because the distance from Test Charge 1 to the \begin{align*}-1\mathrm{C}\end{align*} is only slightly longer than the distance from Test Charge 1 to the 1C charge, the vectors will be similar in length. Once we have determined the relative scale of each vector, we can add them using the parallelogram method. The resultant vector is the electric field at that point.

Finding the electric field at Test Charge 2 will involve all of the same steps. First we must determine which charge Test Charge 2 is closer to. Like Test Charge 1, Test Charge 2 is closer to the 1C charge. However, Test Charge 2 is drastically closer whereas Test Charge 1 was only slightly closer. Therefore, the electric field that Test Charge 2 experiences as a result of the 1C charge will be strong, thus resulting in a longer arrow. The distance between the \begin{align*}-1C\end{align*} and Test Charge 2 is large and therefore the electric field experienced by Test Charge 2 as a result of the \begin{align*}-1C\end{align*} charge will be small. The resultant vectors will look something like this.

## Electric Potential

Like gravity, the electric force can do work and has a potential energy associated with it. But like we use fields to keep track of electromagnetic forces, we use electric potential, or voltage to keep track of electric potential energy. So instead of looking for the potential energy of specific objects, we define it in terms of properties of the space where the objects are.

The electric potential difference, or voltage difference (often just called voltage) between two points (A and B) in the presence of an electric field is defined as the work it would take to move a positive test charge of magnitude 1 from the first point to the second against the electric force provided by the field. For any other charge \begin{align*} q \end{align*}, then, the relationship between potential difference and work will be:

\begin{align*}\Delta V_{AB} &= \frac{W_{AB}}{q} \ \ \ \ \text{[4] Electric Potential} \intertext{Rearranging, we obtain:} \underbrace{W}_{\text{Work}} &= \underbrace{\Delta V_{AB}}_{\text{Potential Difference}} \times \underbrace{q}_{\text{Charge}} \intertext{The potential of electric forces to do work corresponds to electric potential energy:} \Delta U_{E,AB} &= q\Delta V_{AB} \ \ \text{[5] Potential energy change due to voltage change}\end{align*}

The energy that the object gains or losses when traveling through a potential difference is supplied (or absorbed) by the electric field --- there is nothing else there. Therefore, it follows that electric fields contain energy.

To summarize: just as an electric field denotes force per unit charge, so electric potential differences represent potential energy differences per unit charge. A useful mnemonic is to consider a cell phone: the battery has the potential to do work for you, but it needs a charge! Actually, the analogy there is much more rigorous than it at first seems; we'll see why in the chapter on current. Since voltage is a quantity proportional to work it is a scalar, and can be positive or negative.

## Electric Field of a Parallel Plate Capacitor

Suppose we have two parallel metal plates set a distance \begin{align*}d\end{align*} from one another. We place a positive charge on one of the plates and a negative charge on the other. In this configuration, there will be a uniform electric field between the plates pointing from, and normal to, the plate carrying the positive charge. The magnitude of this field is given by

\begin{align*}E = \frac{V}{d}\end{align*}

where \begin{align*}V\end{align*} is the potential difference (voltage) between the two plates.

The amount of charge, \begin{align*}Q\end{align*}, held by each plate is given by

\begin{align*}Q = CV\end{align*}

where again \begin{align*}V\end{align*} is the voltage difference between the plates and \begin{align*}C\end{align*} is the capacitance of the plate configuration. Capacitance can be thought of as the capacity a device has for storing charge . In the parallel plate case the capacitance is given by

\begin{align*}C = \frac{\epsilon_0 A}{d}\end{align*}

where \begin{align*}A\end{align*} is the area of the plates, \begin{align*} d \end{align*} is the distance between the plates, and \begin{align*} \epsilon_0 \end{align*} is the permittivity of free space whose value is \begin{align*} 8.84 \times 10^{-12} C/V \cdot m \end{align*}.

The electric field between the capacitor plates stores energy. The electric potential energy, \begin{align*}U_{\text{C}} \end{align*}, stored in the capacitor is given by

\begin{align*}U_{\text{C}} = \frac{1}{2}CV^2\end{align*}

Where does this energy come from? Recall, that in our preliminary discussion of electric forces we assert that “like charges repel one another”. To build our initial configuration we had to place an excess of positive and negative charges, respectively, on each of the metal plates. Forcing these charges together on the plate had to overcome the mutual repulsion that the charges experience; this takes work. The energy used in moving the charges onto the plates gets stored in the field between the plates. It is in this way that the capacitor can be thought of as an energy storage device. This property will become more important when we study capacitors in the context of electric circuits in the next chapter.

Note: Many home-electronic circuits include capacitors; for this reason, it can be dangerous to mess around with old electronic components, as the capacitors may be charged even if the unit is unplugged. For example, old computer monitors (not flat screens) and TVs have capacitors that hold dangerous amounts of charge hours after the power is turned off.

## More on Electric and Gravitational Potential

There are several differences between our approach to gravity and electricity that could cause confusion. First, with gravity we usually used the concept of “energy”, rather than “energy difference”. Second, we spoke about it in absolute terms, rather than “per unit mass”.

To address the first issue: when we dealt with gravitational potential energy we had to set some reference height \begin{align*} h = 0 \end{align*} where it is equal to \begin{align*} mg \times 0 = 0 \end{align*}. In this sense, we were really talking about potential energy differences rather than absolute levels then also: at any point, we compared the gravitational potential energy of an object to the energy it would have had at the reference level \begin{align*} h = 0 \end{align*}. When we used the formula

\begin{align*}U_g = mg\Delta h\end{align*}

we implicitly set the initial point as the zero: no free lunch! For the same reason, we use the concept of electric potential difference between two points --- or we need to set the potential at some point to 0, and use it as a reference. This is not as easy in this case though; usually a point very far away (“infinitely” far) is considered to have 0 electric potential.

Regarding the second issue: in the chapter on potential energy, we could have gravitational potential difference between two points at different heights as \begin{align*} g\Delta h \end{align*}. This, of course, is the work required to move an object of mass one a height \begin{align*} \Delta h \end{align*} against gravity. To find the work required for any other mass, we would multiply this by its magnitude. In other words,

\begin{align*}\underbrace{W}_{\text{Work}} =\underbrace{m}_{\text{Mass}} \times \underbrace{g\Delta h}_{\text{Potential Difference}}\end{align*}

Which is exactly analogous to the equation above.

## Summary of Relationships

The following table recaps the relationships discussed in this chapter.

Relationship between “per Coulomb” and absolute quantities.
Property of Object. Property of Space. Combine Into:
Charge (Coulombs) Field* (Newtons/Coulomb) Force (Newtons)
Charge (Coulombs) Potential* (Joules/C) Potential Energy (Joules)
• An advanced note: for a certain class of forces called conservative forces e.g., gravity and the electromagnetic force, a specific potential distribution corresponds to a unique field. Conversely, a field corresponds to a unique potential distribution up to an additive constant. Remember though, it's relative potential between points not absolute potential that is physically relevant. In effect the field corresponds to a unique potential. In particular, we see that in the case of conservative forces the scalar potential (one degree of freedom per point) carries all information needed to determine the vector electric field (three degrees of freedom per point. The potential formulation is even more useful than it at first seems.

## Key Concepts

• Electrons have negative charge and protons have positive charge. The magnitude of the charge is the same for both: \begin{align*} e= 1.6 \times 10^{-19} \text{C}\end{align*}.
• In any closed system, electric charge is conserved. The total electric charge of the universe does not change. Therefore, electric charge can only be transferred not lost from one body to another.
• Normally, electric charge is transferred when electrons leave the outer orbits of the atoms of one body (leaving it positively charged) and move to the surface of another body (causing the new surface to gain a negative net charge). In a plasma, the fourth state of matter, all electrons are stripped from the atoms, leaving positively charged ions and free electrons.
• Similarly-charged objects have a repulsive force between them. Oppositely charged objects have an attractive force between them.
• The value of the electric field tells you the force that a charged object would feel if it entered this field. Electric field lines tell you the direction a positive charge would go if it were placed in the field.
• Electric potential is measured in units of Volts (V) thus electric potential is often referred to as voltage. Electric potential is the source of the electric potential energy.
• Positive charges move towards lower electric potential; negative charges move toward higher electric potential. If you are familiar with a contour map then positive charges go 'downhill' and negative charges go 'uphill'.
• Faraday cages consist of a metal box. All of your sensitive electronics are encased in a metal box called a Faraday cage. The Faraday cage protects everything inside from external electric fields. Basically the electrons in the metal box move around to cancel out the electric field, thus preventing it from coming inside the box and thus preventing movement of charge and possible blown out electronic chips. Cars and airplanes, being enclosed in metal, are also Faraday cages and thus the safest place to be in a lightning storm.

## Key Applications

• In problems that ask for excess negative or positive charge, remember that each electron has one unit of the fundamental charge \begin{align*} e = 1.6 \times 10^{-19} \text{C}\end{align*}.
• To find the speed of a particle after it traverses a voltage difference, use the equation for the conservation of energy: \begin{align*} q \Delta V = \tfrac{1}{2} m v^2 \end{align*}
• Force and electric field are vectors. Use your vector math skills (i.e. keep the x and y directions separate) when solving two-dimensional problems.

## Electricity Problem Set

1. After sliding your feet across the rug, you touch the sink faucet and get shocked. Explain what is happening.
2. What is the net charge of the universe? Of your toaster?
3. As you slide your feet along the carpet, you pick up a net charge of \begin{align*}+4 \;\mathrm{mC}\end{align*}. Which of the following is true?
1. You have an excess of \begin{align*}2.5 \times 10^{16}\end{align*} electrons
2. You have an excess of \begin{align*}2.5 \times 10^{19}\end{align*} electrons
3. You have an excess of \begin{align*}2.5 \times 10^{16}\end{align*} protons
4. You have an excess of \begin{align*}2.5 \times 10^{19}\end{align*} protons
4. You rub a glass rod with a piece of fur. If the rod now has a charge of \begin{align*}-0.6\ \mu C\end{align*}, how many electrons have been added to the rod?
1. \begin{align*}3.75 \times 10^{18}\end{align*}
2. \begin{align*}3.75 \times 10^{12}\end{align*}
3. \begin{align*}6000\end{align*}
4. \begin{align*}6.00 \times 10^{12}\end{align*}
5. Not enough information
5. What is the direction of the electric field if an electron initially at rest begins to move in the North direction as a result of the field?
1. North
2. East
3. West
4. South
5. Not enough information

6. Two metal plates have gained excess electrons in differing amounts through the application of rabbit fur. The arrows indicate the direction of the electric field which has resulted. Three electric potential lines, labeled \begin{align*}A, B,\end{align*} and \begin{align*}C\end{align*} are shown. Order them from the greatest electric potential to the least.
1. \begin{align*}A, B, C\end{align*}
2. \begin{align*}C, B, A\end{align*}
3. \begin{align*}B, A, C\end{align*}
4. \begin{align*}B, C, A\end{align*}
5. \begin{align*}A = B = C \ldots\end{align*} they are all at the same potential

7. The diagram to the right shows a negatively charged electron. Order the electric potential lines from greatest to least.
1. \begin{align*}A, B, C\end{align*}
2. \begin{align*}C, B, A\end{align*}
3. \begin{align*}B, A, C\end{align*}
4. \begin{align*}B, C, A\end{align*}
5. \begin{align*}A = B = C \ldots\end{align*} they are all at the same electric potential
8. The three arrows shown here represent the magnitudes of the electric field and the directions at the tail end of each arrow. Consider the distribution of charges which would lead to this arrangement of electric fields. Which of the following is most likely to be the case here?
1. A positive charge is located at point \begin{align*}A\end{align*}
2. A negative charge is located at point \begin{align*}B\end{align*}
3. A positive charge is located at point \begin{align*}B\end{align*} and a negative charge is located at point \begin{align*}C\end{align*}
4. A positive charge is located at point \begin{align*}A\end{align*} and a negative charge is located at point \begin{align*}C\end{align*}
5. Both answers a) and b) are possible
9. Particles \begin{align*}A\end{align*} and \begin{align*}B\end{align*} are both positively charged. The arrows shown indicate the direction of the forces acting on them due to an applied electric field (not shown in the picture). For each, draw in the electric field lines that would best match the observed force.
10. To the right are the electric potential lines for a certain arrangement of charges. Draw the direction of the electric field for all the black dots.
11. A suspended pith ball possessing \begin{align*}+10 \ \mu \mathrm{C}\end{align*} of charge is placed \begin{align*}0.02 \;\mathrm{m}\end{align*} away from a metal plate possessing \begin{align*}-6 \ \mu \mathrm{C}\end{align*} of charge.
1. Are these objects attracted or repulsed?
2. What is the force on the negatively charged object?
3. What is the force on the positively charged object?
12. Calculate the electric field a distance of \begin{align*}4.0 \;\mathrm{mm}\end{align*} away from a \begin{align*}-2.0 \ \mu \mathrm{C}\end{align*} charge. Then, calculate the force on a \begin{align*}-8.0 \ \mu \mathrm{C}\end{align*} charge placed at this point.
13. Consider the hydrogen atom. Does the electron orbit the proton due to the force of gravity or the electric force? Calculate both forces and compare them. (You may need to look up the properties of the hydrogen atom to complete this problem.)
14. As a great magic trick, you will float your little sister in the air using the force of opposing electric charges. If your sister has \begin{align*}40 \;\mathrm{kg}\end{align*} of mass and you wish to float her \begin{align*}0.5 \;\mathrm{m}\end{align*} in the air, how much charge do you need to deposit both on her and on a metal plate directly below her? Assume an equal amount of charge on both the plate and your sister.
15. Copy the arrangement of charges below. Draw the electric field from the \begin{align*}-2 \;\mathrm{C}\end{align*} charge in one color and the electric field from the \begin{align*}+2 \;\mathrm{C}\end{align*} charge in a different color. Be sure to indicate the directions with arrows. Now take the individual electric field vectors, add them together, and draw the resultant vector. This is the electric field created by the two charges together.
16. A proton traveling to the right moves in between the two large plates. A vertical electric field, pointing downwards with magnitude \begin{align*}3.0 \;\mathrm{N/C}\end{align*}, is produced by the plates.
1. What is the direction of the force on the proton?
2. Draw the electric field lines on the diagram.
3. If the electric field is \begin{align*}3.0 \;\mathrm{N/C}\end{align*}, what is the acceleration of the proton in the region of the plates?
4. Pretend the force of gravity doesnot exist; then sketch the path of the proton.
5. We take this whole setup to another planet. If the proton travels straight through the apparatus without deflecting, what is the acceleration of gravity on this planet?
17. A molecule shown by the square object shown below contains an excess of \begin{align*}100\end{align*} electrons. (a) What is the direction of the electric field at point A, \begin{align*}2.0 \times 10^{-9} \;\mathrm{m}\end{align*} away? (b) What is the value of the electric field at point \begin{align*}A\end{align*}? (c) A molecule of charge \begin{align*}8.0 \ \mu \mathrm{C}\end{align*} is placed at point \begin{align*}A\end{align*}. What are the magnitude and direction of the force acting on this molecule?
18. Two negatively charged spheres (one with \begin{align*}-12 \ \mu \mathrm{C}\end{align*}; the other with \begin{align*}-3 \ \mu \mathrm{C}\end{align*}) are \begin{align*}3 \;\mathrm{m}\end{align*} apart. Where could you place an electron so that it will be suspended in space between them with zero net force?

For problems 19, 20, and 21 assume \begin{align*}3-\end{align*}significant digit accuracy in all numbers and coordinates. All charges are positive.

1. Find the direction and magnitude of the force on the charge at the origin (see picture). The object at the origin has a charge of \begin{align*}8 \ \mu \mathrm{C}\end{align*}, the object at coordinates \begin{align*}(-2 \;\mathrm{m}, \ 0)\end{align*} has a charge of \begin{align*}12 \ \mu \mathrm{C}\end{align*}, and the object at coordinates \begin{align*}(0, -4 \;\mathrm{m})\end{align*} has a charge of \begin{align*}44 \ \mu \mathrm{C}\end{align*}. All distance units are in meters.
2. A \begin{align*}2 \;\mathrm{C}\end{align*} charge is located at the origin and a \begin{align*}7 \;\mathrm{C}\end{align*} charge is located at \begin{align*}(0, 6 \;\mathrm{m})\end{align*}. Find the electric field at the coordinate \begin{align*}(10 \;\mathrm{m}, 0)\end{align*}. It may help to draw a sketch.
3. A metal sphere with a net charge of \begin{align*}+5\ \mu \mathrm{C}\end{align*} and a mass of \begin{align*}400 \;\mathrm{g}\end{align*} is placed at the origin and held fixed there.
1. Find the electric potential at the coordinate \begin{align*}(6 \;\mathrm{m}, 0)\end{align*}.
2. If another metal sphere of \begin{align*}-3 \ \mu \mathrm{C}\end{align*} charge and mass of \begin{align*}20 \;\mathrm{g}\end{align*} is placed at the coordinate \begin{align*}(6 \;\mathrm{m}, 0)\end{align*} and left free to move, what will its speed be just before it collides with the metal sphere at the origin?

4. Collisions of electrons with the surface of your television set give rise to the images you see. How are the electrons accelerated to high speed? Consider the following: two metal plates (The right hand one has small holes allow electrons to pass through to the surface of the screen.), separated by \begin{align*}30 \;\mathrm{cm}\end{align*}, have a uniform electric field between them of \begin{align*}400 \;\mathrm{N/C}\end{align*}.
1. Find the force on an electron located at a point midway between the plates
2. Find the voltage difference between the two plates
3. Find the change in electric potential energy of the electron when it travels from the back plate to the front plate
4. Find the speed of the electron just before striking the front plate (the screen of your TV)

5. Two pith balls of equal and like charges are repulsed from each other as shown in the figure below. They both have a mass of \begin{align*}2 \;\mathrm{g}\end{align*} and are separated by \begin{align*}30^\circ\end{align*}. One is hanging freely from a \begin{align*}0.5 \;\mathrm{m}\end{align*} string, while the other, also hanging from a \begin{align*}0.5\;\mathrm{m}\end{align*} string, is stuck like putty to the wall.
1. Draw the free body diagram for the hanging pith ball
2. Find the distance between the leftmost pith ball and the wall (this will involve working a geometry problem)
3. Find the tension in the string (Hint: use \begin{align*}y-\end{align*}direction force balance)
4. Find the amount of charge on the pith balls (Hint: use \begin{align*}x-\end{align*}direction force balance)

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