## Test Your Knowledge

### Charge

1. What force does a 1 mC charged particle feel when placed 1 m away from a 2 mC charged particle? Is the force attractive or repulsive?

2. How much does the force increase if the 2 mC charge in the previous question is doubled to 4 mC? What if, instead of doubling the charge, we halved the original distance to 50 cm?

3. If you push on a 1 μC charge with a force of 10 N, how close can you get it to a stationary 5 μC charge?

4. What is the force on a 1 mC charge that sits between a -2 mC charge (1.5 m to its left) and a 0.5 mC charge (0.5 m to its right)?

5. Imagine an equilateral triangle with 1 m long sides and a charge at each vertex, two positive (1 mC and 2 mC) and one negative (-1 mC).

What is the magnitude and direction of the force on the -1 mC charge?

### Electric Fields

1. Draw the electric field lines for a -1 C point charge.

2. What is the net electric field created 10 m away from a +1 C and a -0.5 C charge held very close together?

3. What force does an electron feel in a uniform electric field of 0.5 N/C?

4. A point charge makes an E-field of 10 N/C through a sphere of radius 2 m. What is the electric flux created by it?

5. What is the electric field a distance *r* away from an infinitely long charged rod with a linear charge density of λ C/m?

### Electric Potential

1. Draw the equipotential lines for a constant electric field between two charged plates.

2. What is the electric potential at a point 1 m to the right of a 1 nC charge and 1.5 m to the right of a 2 nC charge?

3. What is the potential difference between a positively charged plate and a negatively charged plate that create a constant electric field *E* in a region of width *d* between them?

4. A test charge *q* is placed a distance *r*_{a} away from a second charge *Q*. What is the work that must be done to move the test charge to a new point *r*_{b}?

5. A 2 mC charge is brought to within 1 m of a 4 mC charge and then released. When the 2 mC charge, which weighs 3 kg, is a very long distance away from the 4 mC charge, how fast will it be traveling?

### Basic Circuit Elements

1. A common semiconductor manufacturing technique is something called electron beam lithography, where a stream of electrons is used to remove designated sections of material and create a functional circuit in a very small area. If an e-beam machine can create 1 nA of current, how many electrons are hitting the surface of its target with each second?

2. A tennis ball launcher fires one tennis ball every two seconds towards a player practicing her forehand. If the contact between the fibers of the ball and the belt of the launcher charges each ball with 0.1 mC of static electricity, what is the electric current created by the stream of tennis balls?

3. What is the voltage drop across a 1000 Ω resistor with a 50 mA current flowing through it?

4. What is the current through a wire with an internal resistance of 10 Ω when it is attached to a 1.5 V battery?

5. What is the resistance of a piece of metal that allows 100 mA of current to pass through it when it's attached to a 7.2 V battery?

### Solving Circuit Questions

1. We worked out the parallel case, but now let's tackle the series combination of two resistors. If *V* = 5 V, *R*_{1} = 100Ω, and *R*_{2} = 200Ω, what is the voltage drop across each resistor? What is the current through each resistor? What is the current provided by the battery?

2. What is the current provided by the battery in this circuit?

3. How much power is dissipated in a 1000 Ω resistor when 10 V are applied to it?

4. What is the equivalent resistance of this network of resistors? Each individual resistor is 100 Ω.

5. What is the power dissipated in resistor *R*_{4} in this circuit?

### Magnetic Fields

1. What is the magnetic field created 1 mm away from a wire carrying 10 A of current?

2. A wire carries current to the right. What direction is the magnetic field directly above the wire? Below it?

3. A 24 V battery is attached to a long, straight wire. When the circuit is connected, what is the maximum resistance the wire could have in order to create a minimum magnetic field of 1 mT a distance of 1 mm from the wire?

4. A solenoid is a coil of wire that creates a magnetic field very similar to that of a bar magnet. If a solenoid *L* meters long has *N* turns (*N* individual loops of wire making up the coil), what is the magnetic field created inside it when a current *I* flows through the solenoid?

5. Which has more energy: the AM radio waves received in your car stereo, or the FM radio waves?

### Magnetic Forces

1. A +1 C charge moves at 5 m/s to the right in a magnetic field of 0.5 T into the screen.

What force does the charge feel? What direction does the force act in?

2. A square loop of wire sitting in a magnetic field out of the screen has a current traveling clockwise around it. Does the Lorentz force try to expand or compress the loop?

3. A *velocity selector* is a spectroscopy tool used to choose ions (charged particles) with specific velocities. It accomplishes this by accelerating particles through perpendicular electric and magnetic fields, such that the ions are pulled to the left by an electric field and pulled to the right by a magnetic field. Only ions with exactly the right speed will travel through the exit aperture, which is barely bigger than the accelerated ions and is aligned perfectly with the entrance point—any deflection left or right will send the ions crashing into the velocity selector's wall instead. What velocity, in terms of the strength of the electric field ( *E*) and magnetic field (*B*), will let ions exit the selector without deflection?

4. If an electron is moving to the right, in a magnetic field that is pointed into the screen, what direction is the force?

5. A 0.2 kg particle with 2 C of charge is moving at 15 m/s to the right through a magnetic field *B*. How strong and in what direction does *B* need to be to cancel out the force of gravity on the charged particle?

### Electromagnetic Induction

1. What is the magnetic flux through a circular loop of wire with a 3 m diameter placed in a constant magnetic field of 0.5 T?

2. A solenoid (coil) consisting of 100 loops of wire generates a voltage of -11 V when an electromagnet near it is turned on, a process that takes 100 ms. What is the change of flux seen by the solenoid?

3. When you drop a magnet down a copper pipe (north end pointing down), which way must the current induced in the pipe initially flow to slow the magnet's fall?

4. A square loop of wire with 3 cm sides rests on a table. When a large magnet is brought near it over the course of 5 ms, a magnetic field appears pointing straight down into the table, reaching 0.4 T at its peak. If the resistance of the loop of wire is 10 Ω, what current appears in the loop? What direction does it flow?

5. A square loop of wire with 5 cm sides moves to the right on a table at 1 m/s. Eventually it enters a region of the table with a constant magnetic field of 0.5 T pointing straight up. What emf is generated in the wire?

### Advanced Circuit Elements

1. What current is provided by a battery with \mathcal{E} = 10 V and internal resistance *r* = 5Ω when it's connected to a 100 Ω resistor?

2. A simple parallel plate capacitor consists of two 2 cm by 3 cm squares of aluminum separated by a distance of 1 mm. How many electrons can be stored on the negatively charged plate when 12 V is applied to the capacitor?

3. In an *RC circuit*, a battery with ideal voltage *V* is attached in series to a resistor with resistance *R* and a capacitor with capacitance *C*. What is the charge *q* present on the capacitor plates as a function of *V*, *R*, *C*, and *I* (the current instantaneously provided by the battery)?

4. How much energy is stored in this capacitor network? Each capacitor is 10 μF and the network is attached to a 7.2 V (ideal) battery.

5. How much energy is stored in this inductor network? Each inductor is 1 mH and the network is attached to a current source that provides 2 A of current.

### Answers

### Charge

1. *What force does a 1 mC charged particle feel when placed 1 m away from a 2 mC charged particle? Is the force attractive or repulsive?*

Using Coulomb's Law: .Because both charges are positive, this force is repulsive—as shown by the positive answer for *F*_{e}.

2. *How much does the force increase if the 2 mC charge in the previous question is doubled to 4 mC? What if, instead of doubling the charge, we halved the original distance to 50 cm?*

When we double the charge, we get .So *F*_{e} is quadrupled.

3. *If you push on a 1 μC charge with a force of 10 N, how close can you get it to a stationary 5 μC charge?*

If you can provide 10 N of force, you cannot push the charge any closer than the point at which the repulsive Coulomb force is also 10 N. This occurs at .

4. *What is the force on a 1 mC charge that sits between a -2 mC charge (1.5 m to its left) and a 0.5 mC charge (0.5 m to its right)?*

Using Coulomb's Law and the principle of superposition: .The force exerted on the 1 mC particle by the -2 mC particle is .This is an attractive force that *pulls* the particle to the left. The force exerted on the 1 mC particle by the 0.5 mC particle is .This is a repulsive force that *pushes* our particle to the left. Therefore the total force on our 1 mC particle is to the left.

5. *Imagine an equilateral triangle with 1 m long sides and a charge at each vertex, two positive (1 mC and 2 mC) and one negative (-1 mC).*

*What is the magnitude and direction of the force on the -1 mC charge?*

The force on the -1 mC charge from the 1 mC charge is .Since the force is negative (attractive), this points up and to the left along the line between the -1 and 1 mC charges. The force on the -1 mC charge from the 2 mC charge is .Since the force is negative (attractive), this points up and to the right along the line between the -1 and 2 mC charges. Now, we can use the principle of superposition to find the net force on the -1 mC charge. Here are our two forces, broken into *x* and *y* components:

*F*_{e,1,x} is given by *F*_{e,1}cos(60º) and *F*_{e,1,y} is given by *F*_{e,1}sin(60º). We can find the components of *F*_{e,2} similarly. For our net force we then have *||F*_{e,tot,x}|| = *||F*_{e,2,x}*||* – *||F*_{e,1,x}*||* = 18,000cos(60º) – 9000cos(60º) = 4500 N and *||F*_{e,tot,y}*||* = *||F*_{e,2,y}*||* + *||F*_{e,1,y}*||* = 18,000sin(60º) + 9000sin(60º) = 23,383 N. Combining these back into a magnitude and direction gives and up to the right

### Electric Fields

1. *Draw the field lines for a -1 C point charge.*

Remember the arrows on the field lines point in the direction a +1 C charge would travel in the field.

2. *What is the net electric field created 10 m away from a +1 C and a -0.5 C charge held very close together?*

Since charges obey superposition, we can sum the field each charge would create on its own to get a net electric field. The positive charge creates an E-field of . The negative charge's field is . Therefore the net electric field is *E*_{tot} = E_{+} + *E*_{-} = +4.5 × 10^{7} N/C.

3. *What force does an electron feel in a uniform electric field of 0.5 N/C?*

The force on an electron is given by *F*_{e} = *qE* = (1.6 × 10^{-19} C) × (0.5 N/C) = 0.8 × 10^{-19} N. This may seem small, but since an electron has a mass of 9.1 × 10^{-31} kg, this modest field will accelerate the electron at almost nine billion *g*'s.

4. *What is the electric flux created by a point charge that makes an E-field of 10 N/C through a sphere of radius 2m centered at the point charge?*

Φ_{E} = *EA* = *E*(4π *r*^{2}) = (10 N/C) × (4π × (2 m)^{2}) = 502.7 Nm^{2}/C.

5. *What is the electric field a distance r away from an infinitely long charged rod with a linear charge density of λ C/m?*

Gauss' Law to the rescue here. Let's look at a length of rod, *l*, and draw a Gaussian cylinder around it:

Gauss' Law says that . The total charge *q* enclosed in our section of length *l* will be q = λ *l*. Since the electric field radiates outward from the rod, the surface area that flux passes through is equal to the sidewall of the Gaussian cylinder (no field lines pass through the ends): *A* = 2π*rl*. Therefore, we have , giving us an electric field of at every point along the rod—there's no dependence on our made up variable *l*, which is a good sign we've stumbled onto the correct answer.

### Electric Potential

1. *Draw the equipotential lines for a constant electric field between two charged plates.*

*Answer*:

2. *What is the electric potential at a point 1 m to the right of a 1 nC charge and 1.5 m to the right of a 2 nC charge?*

The potential from the 2 nC charge is . The potential from the 1 nC charge is . The net potential seen at the point in question is then *V*_{tot} = V _{1} + V _{2} = 21 V.

3. *What is the potential difference between a positively charged plate and a negatively charged plate that create a constant electric field E in a region of width d between them?*

The potential difference, Δ*V*, must be equal to the work done per unit charge to move a positively charged particle *q* from the negative plate to the positive plate (against the Coulomb force).

The work done per unit charge is , where . Therefore, the potential difference is Δ*V* = *Ed*.

4. *A test charge q is placed a distance r_{a} away from a second charge Q. What is the work that must be done to move the test charge to a new point r_{b}?*

The electric potential energy of charge *q* at point *r*_{a} is ,and the potential energy at point *r*_{b} is .The work done moving *q* from *r*_{a} to *r*_{b} is then

5. *A 2 mC charge is brought to within 1 m of a 4 mC charge and then released. When the 2 mC charge, which weighs 3 kg, is a very long distance away from the 4 mC charge, how fast will it be traveling?*

When it's brought to a point 1 m away from a 4 mC charge, the 2 mC charge has an electric potential energy of . When the 2 mC charge is very far away from the 4 mC charge, *r* is very large, and so *U*_{e} is negligible. This means all of the potential energy has been turned into kinetic energy: . The 2 mC charge's final speed is therefore .

### Basic Circuit Elements

1. *A common semiconductor manufacturing technique is something called electron beam lithography, where a stream of electrons is used to remove designated sections of material and create a functional circuit in a very small area. If an e-beam machine can create 1 nA of current, how many electrons are hitting the surface of its target with each second?*

The current created by the e-beam is , where *e* is the elementary charge and *n* is the number of electrons in the beam. In 1 s, the number of electrons that hit the target is electrons.

2. *A tennis ball launcher fires one tennis ball every two seconds towards a player practicing her forehand. If the contact between the fibers of the ball and the belt of the launcher charges each ball with 0.1 mC of static electricity, what is the electric current created by the stream of tennis balls?*

3. *What is the voltage drop across a 1000 Ω resistor with a 50 mA current flowing through it?*

*V* = *IR* = (0.05 A) × (1000Ω) = 50 V

4. *What is the current through a wire with an internal resistance of 10 Ω when it is attached to a 1.5 V battery?*

5. *What is the resistance of a piece of metal that allows 100 mA of current to pass through it when it's attached to a 7.2 V battery?*

### Solving Circuit Questions

1. *We worked out the parallel case, but now let's tackle the series combination of two resistors. If V = 5 V, R_{1} = 100Ω, and R_{2} = 200Ω, what is the voltage drop across each resistor? What is the current through each resistor? What is the current provided by the battery?*

In this case, the easy value to solve for is current. The two resistors can be replaced with an equivalent resistor *R*_{s} = *R*_{1} + *R*_{2} = 300Ω. The current through both resistors is the same, and equal to the current supplied by the battery: . The voltage drop across each resistor is then *V*_{1} = *IR*_{1} = (0.0167 A) × (100Ω) = 1.67 V and V _{2} = IR _{2} = (0.0167 A) × (200Ω) = 3.33 V. By KVL, we know the net voltage change around the loop must be zero—and it is: the 1.67 V and 3.33 V drops across *R*_{1} and *R*_{2} sum to exactly the 5 V added to the circuit by the battery.

2. *What is the current provided by the battery in this circuit?*

To find the current provided by the battery, *I*_{tot}, we must replace the three resistors witha single equivalent resistor. Combining the two parallel resistors gives . The total equivalent resistance is a series combination of *R*_{1} and . Therefore the current provided by the battery is . Now, for the final calculation: .

3. *How much power is dissipated in a 1000 Ω resistor when 10 V are applied to it?*

Combining our equation for resistive power dissipation (*P* = *I*^{2}*R*) and Ohm's Law (*V* = *IR*) gives us .

4. *What is the equivalent resistance of this network of resistors? Each individual resistor is 100 Ω.*

Start from the inside and work your way out. *R*_{3} and *R*_{4} combine in series, giving us *R*_{s} = *R*_{3} + *R*_{4} = 200Ω. *R*_{2}, *R*_{s}, and *R*_{5} are all in parallel, and so combine to give . Finally, *R*_{1} and *R*_{||} combine in series, giving us a total equivalent resistance of *R*_{eq} = *R*_{1} + *R*_{||} = 140Ω.

5. *What is the power dissipated in resistor R_{4} in this circuit?*

Strap yourself in. To find the power dissipated in *R*_{4}, we must either find the current through it (*P*_{4} = *I*_{4}^{2}*R*_{4}) or the voltage drop across it (we can combine our equation for power and Ohm's Law to get . Because *R*_{4} is in series with *R*_{5}, we will never know the voltage drop across it unless we also know the current passing through it, so this makes our decision easy—let's find the current *I*_{4} through *R*_{4}. To do this, we need to know the total current provided by the battery so that we can find the current that flows into the branch with *R*_{2}, *R*_{3}, *R*_{4}, and *R*_{5}, so that we can find the current that flows into *R*_{4}. No problem, right?

Start by finding an equivalent resistor to replace this whole mess. *R*_{4} and *R*_{5} combine in series to give *R*_{s(4,5)} = (50Ω) + (100Ω) = 150Ω. *R*_{s(4,5)} combines in parallel with *R*_{3} to give . *R*_{||(3,4,5)} combines in series with *R*_{2} to give *R*_{s(2,3,4,5)} = (200Ω) + (100Ω) = 300Ω. Finally, *R*_{s(2,3,4,5)} combines in parallel with *R*_{1} to give our overall equivalent resistance of. Then we have . Cool. That current is going to flow into the first node, and some of it will go to *I*_{1}, some to *I*_{2} (which in turn will split into *I*_{3} and *I*_{4}). Here's where KVL and KCL come into play. We know the net voltage change in any current path is zero. There are three possible paths in this circuit: through *R*_{1} ; through *R*_{2} and *R*_{3}; or through *R*_{2}, *R*_{4}, and *R*_{5}. Each path must have a voltage drop of 12 V to balance out the 12 V added by the battery. That means the current in *R*_{1} is . By KCL, the current into that first node must equal the current out, so .

Now we have a much smaller problem: four resistors (*R*_{2}, *R*_{3}, *R*_{4}, and *R*_{5}), a 12 V drop, and 40 mA of current. All 40 mA will pass through *R*_{2}, creating a voltage drop of *V* _{2} = I _{2}R _{2} = (0.04 A) × (200Ω) = 8 V. And now our problem's even smaller—there's a (12 V) – (8 V) = 4 V drop across the parallel combination of *R*_{3}, *R*_{4}, and *R*_{5}. We know the current through *R*_{4} and *R*_{5} will be the same, since they're in series, and we can find that current using their equivalent series resistance: .

This is, finally, the current number we need. The power dissipated in *R*_{4} is *P*_{4} = *I*_{4}^{2}*R*_{4} = (0.0267 A)^{2} × (50Ω) = 35.6 mW. Phew.

### Magnetic Fields

1. *What is the magnetic field created 1 mm away from a wire carrying 10 A of current?*

2. *A wire carries current to the right. What direction is the magnetic field directly above the wire? Below it?*

Using the right hand rule with your thumb pointed to the right, your fingers curl towards you. Above the wire, the field comes out of the screen towards you. But below the wire, the field goes back into the screen, away from you.

3. *A 24 V battery is attached to a long, straight wire. When the circuit is connected, what is the maximum resistance the wire could have in order to create a minimum magnetic field of 1 mT a distance of 1 mm from the wire?*

The current needed to create such a magnetic field is given by . To create a 5 A current from a 24 V battery, the wire's resistance can be no larger than .

4. *A solenoid is a coil of wire that creates a magnetic field very similar to that of a bar magnet. If a solenoid L meters long has N turns (N individual loops of wire making up the coil), what is the magnetic field created inside it when a current I flows through the solenoid?*

First, we must draw our Ampèrian loop.

The best loop to choose—one with sides parallel to the field in the solenoid—is a rectangle, with one end at the middle of the coil and one far enough away that the magnetic field is negligible. The *Bl* in Ampère's Law is the product of the length of our coil parallel to the magnetic field and the magnetic field at that point; with a rectangle like this it would be *B*_{in}*L* + *B*_{out}*L*.

Since *B*_{out} at the top of our rectangle is zero, only the bottom side of the rectangle will contribute to Ampère's Law. Therefore we have *B*_{in}*L* = μ _{0}*NI*, since the total current through the solenoid is equal to the current in one loop times the number of loops. This gives a final answer for the magnetic field inside a solenoid of .

5. *Which has more energy: the AM radio waves received in your car stereo, or the FM radio waves?*

FM waves have a higher frequency (10^{8} Hz for FM versus 10^{6} for AM) and are therefore higher energy waves.

### Magnetic Forces

1. *A +1 C charge moves at 5 m/s to the right in a magnetic field of 0.5 T into the screen.*

*What force does the charge feel? What direction does the force act in?*

*F*_{b} = *qvB *sin θ = (1 C) × (5 m/s) × (0.5 T) × sin(90º) = 2.5 N. The right hand rule tells us this force is exerted upwards (index finger points right, middle finger points into the screen).

2. *A square loop of wire sitting in a magnetic field out of the screen has a current traveling clockwise around it. Does the Lorentz force try to expand or compress the loop?*

The right hand rule shows us the magnetic field is trying to compress the loop—the force on each side of the square wire points inwards towards the square's center.

3. *A velocity selector is a spectroscopy tool used to choose ions (charged particles) with specific velocities. It accomplishes this by accelerating particles through perpendicular electric and magnetic fields, such that the ions are pulled to the left by an electric field and pulled to the right by a magnetic field. Only ions with exactly the right speed will travel through the exit aperture, which is barely bigger than the accelerated ions and is aligned perfectly with the entrance point—any deflection left or right will send the ions crashing into the velocity selector's wall instead. What velocity, in terms of the strength of the electric field (E) and magnetic field (B), will let ions exit the selector without deflection?*

The force on a charged particle moving perpendicular to an *E* and *B* field that push it in opposite directions is *F* = *qE – qvB*. If the particle isn't going to be deflected at all, the net force *F* must be 0, so we have . Any ion moving at a different speed will not be able to exit the selector.

4. *If an electron is moving to the right, in a magnetic field that is pointed into the screen, what direction is the force?*

The electron moves to the right, but in the right hand rule we must always point our first finger in the direction *positive* charge is moving. In this case, that's effectively the left. Pointing our first finger left and our second finger into the screen gives a net downward force. This is consistent with the formula *F*_{b} = *qvB *sin θ = -*evB*.

5. *A 0.2 kg particle with 2 C of charge is moving at 15 m/s to the right through a magnetic field B. How strong and in what direction does B need to be to cancel out the force of gravity on the charged particle?*

Gravity will pull the particle down, so the magnetic force must push it up to counter. Putting our first finger to the right in the direction the particle moves and our thumb straight up leaves our middle finger pointing into the screen. That is *B*'s direction. The force on the particle from gravity is *F*_{g} = *mg* and we know this must equal the force on the particle from the magnetic field, *F*_{b} = *qvB*sinθ. Therefore we have into the screen.

### Electromagnetic Induction

1. *What is the magnetic flux through a circular loop of wire with a 3 m diameter placed in a constant magnetic field of 0.5 T?*

2. *A solenoid (coil) consisting of 100 loops of wire generates a voltage of -11 V when an electromagnet near it is turned on, a process that takes 100 ms. What is the change of flux seen by the solenoid?*

3. *When you drop a magnet down a copper pipe (north end pointing down), which way must the current induced in the pipe initially flow to slow the magnet's fall?*

The magnetic field created by the current in the pipe must point upwards to oppose the magnet's field, which is moving downwards. The right hand rule shows that the current in the pipe must flow around it counterclockwise to create such a field.

4. *A square loop of wire with 3 cm sides rests on a table. When a large magnet is brought near it over the course of 5 ms, a magnetic field appears pointing straight down into the table, reaching 0.4 T at its peak. If the resistance of the loop of wire is 10 Ω, what current appears in the loop? What direction does it flow?*

,where ΔΦ_{B} = (Δ*B*)*A* because *A* is held constant while the magnet is brought closer, changing *B*. *N = 1 because there is only one loop of wire, not a stacked coil. Therefore the magnitude of the emf created is ||\mathcal{E}|| = 72 mV, and the current that appears is . The negative sign in Faraday's Law means this current opposes the change in magnetic field; since the magnetic field created by the magnet points into the table, the field generated by the current in the loop of wire must point upwards, away from the table. The right hand rule (rule #1) shows that an upward field will be created by a counterclockwise current.*

5. *A square loop of wire with 5 cm sides moves to the right on a table at 1 m/s. Eventually it enters a region of the table with a constant magnetic field of 0.5 T pointing straight up. What emf is generated in the wire?*

, where because the field is held constant while the loop moves in, changing the area of the loop that experiences a magnetic field from 0 to *A*.

*N* = 1 since there is only one loop, and to find Δ *t* we use the kinematics formula *s* = *v*Δ *t*, where *s* is the length of the square's side—when the square loop enters the magnetic field, the flux changes until it is fully inside the field, at which point Δ *A* drops to zero (when fully within the field, the whole loop sees exactly the same *B*, and so nothing is changing). We have .

### Advanced Circuit Elements

1. *What current is provided by a battery with E = 10 V and internal resistance r = 5Ω when it's connected to a 100 Ω resistor?*

KVL gives tells us the net voltage change around the loop is zero, so the voltage added by the battery must equal that dissipated in its internal resistance and the external resistor:

2. *A simple parallel plate capacitor consists of two 2 cm by 3 cm squares of aluminum separated by a distance of 1 mm. How many electrons can be stored on the negatively charged plate when 12 V is applied to the capacitor?*

The capacitance of the parallel plate capacitor is , where *n* is the number of electrons that are stored on the negative plate. This becomes .

3. *In an RC circuit, a battery with ideal voltage V is attached in series to a resistor with resistance R and a capacitor with capacitance C. What is the charge q present on the capacitor plates as a function of V, R, C, and I (the current instantaneously provided by the battery)?*

Applying KVL says the voltage added by the battery must equal the sum of voltage drop across the resistor and the voltage built up on the capacitor: . Rearranging, q = C(V – IR).

4. *How much energy is stored in this capacitor network? Each capacitor is 10 μF and the network is attached to a 7.2 V (ideal) battery.*

The energy stored in a network of capacitors must be the same as the energy stored in one equivalent capacitor, so first we have to find the equivalent capacitance of the network. *C*_{2} and *C*_{3} combine in parallel to give *C*_{||} = *C*_{2} + *C*_{3} = 20μ F. *C*_{||} then combines in series with *C*_{1} to give an overall equivalent capacitance of .The energy stored in the network when 7.2 V is applied is therefore .

5. *How much energy is stored in this inductor network? Each inductor is 1 mH and the network is attached to a current source that provides 2 A of current.*

First, we have to find the equivalent inductance of the network. *L*_{2} and *L*_{3} combine in parallel to give *L*_{||} then combines in series with *L*_{1} to give an overall equivalent inductance of *L*_{eq} = L_{1} + *L*_{||} = 1.5 mH. The energy stored in the network when 2 A are run through it is therefore .