Electricity and Magnetism Practice

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1.1Force Between Two Point Charges

Exam I | Problem 1.1 |

Coulomb's law, Point charges

Two point charges, $q_1 = +3\,\mu\text{C}$ and $q_2 = -2\,\mu\text{C}$, are separated by $0.50\ \text{m}$.

Find the magnitude of the force between them and state whether the force is attractive or repulsive.

Solution

Use Coulomb's law:

F = k\frac{|q_1q_2|}{r^2}

Substitute the values:

F = 9.0\times 10^9 \frac{(3\times 10^{-6})(2\times 10^{-6})}{(0.50)^2}

F = 0.216\ \text{N}

Because the charges have opposite signs, the force is attractive.

1.2Electric Field of a Point Charge

Exam I | Problem 1.2 |

A charge of $+5.0\ \text{nC}$ is located in space.

What is the magnitude of the electric field $0.20\ \text{m}$ away from the charge, and what is its direction?

Solution

For a point charge,

E = k\frac{|q|}{r^2}

Substitute:

E = 9.0\times 10^9 \frac{5.0\times 10^{-9}}{(0.20)^2}

E = 1125\ \text{N/C}

Since the charge is positive, the field points away from the charge.

1.3Linear Charge Density

Exam I | Problem 1.3 |

A uniform charge of $12\ \mu\text{C}$ is spread along a thin wire of length $3.0\ \text{m}$.

What is the linear charge density $\lambda$?

Solution

Linear charge density is

\lambda = \frac{dq}{dl}

For a uniform wire, this becomes

\lambda = \frac{Q}{L}

\lambda = \frac{12\ \mu\text{C}}{3.0\ \text{m}} = 4.0\ \mu\text{C/m}

1.4Electric Flux Through a Flat Surface

Exam I | Problem 1.4 |

Electric flux

A uniform electric field of magnitude $200\ \text{N/C}$ passes through a flat surface of area $0.15\ \text{m}^2$.

The angle between the field and the surface normal is $30^\circ$.

What is the electric flux?

Solution

For a flat surface in a uniform field,

\Phi_E = EA\cos\theta

Substitute the values:

\Phi_E = (200)(0.15)\cos 30^\circ

\Phi_E \approx 26\ \text{N}\cdot\text{m}^2/\text{C}

1.5Charge Enclosed by a Gaussian Surface

Exam I | Problem 1.5 |

Gauss's law, Electric flux

A closed surface has electric flux

\Phi_E = 5.0\times 10^3\ \text{N}\cdot\text{m}^2/\text{C}.

What charge is enclosed by the surface?

Solution

Gauss's law says

\oint \mathbf{E}\cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}

Q_{\text{enc}} = \varepsilon_0\Phi_E

Substitute:

Q_{\text{enc}} = (8.854\times 10^{-12})(5.0\times 10^3)

Q_{\text{enc}} = 4.4\times 10^{-8}\ \text{C}

1.6Potential from Two Point Charges

Exam I | Problem 1.6 |

At a point in space, a $+4.0\ \text{nC}$ charge is $0.30\ \text{m}$ away and a $-1.0\ \text{nC}$ charge is $0.20\ \text{m}$ away.

What is the electric potential at that point?

Solution

Electric potential adds as a scalar:

V = \sum_i k\frac{q_i}{r_i}

Compute each contribution:

V_1 = 9.0\times 10^9 \frac{4.0\times 10^{-9}}{0.30} = 120\ \text{V}

V_2 = 9.0\times 10^9 \frac{-1.0\times 10^{-9}}{0.20} = -45\ \text{V}

V = 120 - 45 = 75\ \text{V}

1.7Change in Potential Energy

Exam I | Problem 1.7 |

A charge of $+3.0\ \mu\text{C}$ moves from a point at $20\ \text{V}$ to a point at $-10\ \text{V}$.

What is the change in its electric potential energy?

Solution

Use

\Delta U = q\Delta V

with

\Delta V = V_f - V_i = -10 - 20 = -30\ \text{V}

Then

\Delta U = (3.0\times 10^{-6})(-30)

\Delta U = -9.0\times 10^{-5}\ \text{J}

The potential energy decreases.

1.8Dielectric Constant from Capacitance

Exam I | Problem 1.8 |

A capacitor has capacitance $2.0\ \mu\text{F}$ in vacuum. After a dielectric is inserted, its capacitance becomes $6.0\ \mu\text{F}$.

What is the dielectric constant $\kappa$?

Solution

With a dielectric,

C = \kappa \varepsilon_0 \frac{A}{d}

So the capacitance increases by a factor of $\kappa$:

\kappa = \frac{6.0}{2.0} = 3.0

1.9Resistance of a Uniform Wire

Exam I | Problem 1.9 |

A wire has resistivity $2.0\times 10^{-6}\ \Omega\cdot\text{m}$, length $4.0\ \text{m}$, and cross-sectional area $2.0\times 10^{-6}\ \text{m}^2$.

Find its resistance, and then find the current if a $12\ \text{V}$ potential difference is applied across it.

Solution

For a uniform conductor,

R = \rho\frac{L}{A}

Substitute:

R = (2.0\times 10^{-6})\frac{4.0}{2.0\times 10^{-6}} = 4.0\ \Omega

Now use Ohm's law:

$$ V = IR $$

I = \frac{V}{R} = \frac{12}{4.0} = 3.0\ \text{A}

1.10Magnetic Force on a Current-Carrying Wire

Exam I | Problem 1.10 |

A straight wire segment of length $0.30\ \text{m}$ carries a current of $5.0\ \text{A}$. It is placed in a uniform magnetic field of magnitude $0.40\ \text{T}$, and the wire is perpendicular to the field.

What is the magnitude of the magnetic force on the wire?

Solution

For a straight wire in a uniform field,

F = ILB\sin\theta

Here $\theta = 90^\circ$, so

$$ F = (5.0)(0.30)(0.40) $$

F = 0.60\ \text{N}

Difficulty

Integrated

2.1Net Field from Two Point Charges on a Line

Exam II | Problem 2.1 |

Two charges lie on a line: a $+6.0\ \text{nC}$ charge on the left and a $-6.0\ \text{nC}$ charge on the right. A point is exactly halfway between them, $0.10\ \text{m}$ from each charge.

What is the net electric field at the midpoint?

Solution

Each charge produces a field of magnitude

E = k\frac{|q|}{r^2}

So for one charge,

E_1 = 9.0\times 10^9 \frac{6.0\times 10^{-9}}{(0.10)^2} = 5400\ \text{N/C}

At the midpoint, the field from the positive charge points to the right, and the field from the negative charge also points to the right.

So the fields add:

E_{\text{net}} = 2(5400) = 1.08\times 10^4\ \text{N/C}

The direction is to the right, from the positive charge toward the negative charge.

2.2Gauss's Law for an Infinite Sheet

Exam II | Problem 2.2 |

Gauss's law, Electric field

An infinite sheet of charge has surface charge density

\sigma = 3.5\ \mu\text{C/m}^2.

What is the electric field magnitude on either side of the sheet, and what is the direction?

Solution

For an infinite sheet,

E = \frac{\sigma}{2\varepsilon_0}

Substitute:

E = \frac{3.5\times 10^{-6}}{2(8.854\times 10^{-12})}

E \approx 2.0\times 10^5\ \text{N/C}

Because the sheet is positively charged, the field points away from the sheet on both sides.

2.3Work from a Potential Difference

Exam II | Problem 2.3 |

A $3.0\ \mu\text{C}$ charge moves from a point at $20\ \text{V}$ to a point at $-10\ \text{V}$.

How much work does the electric field do on the charge?

Solution

The work done by the field is

W_{\text{field}} = -\Delta U = -q\Delta V

Here,

\Delta V = -10 - 20 = -30\ \text{V}

W_{\text{field}} = -(3.0\times 10^{-6})(-30)

W_{\text{field}} = 9.0\times 10^{-5}\ \text{J}

2.4Capacitors in Series

Exam II | Problem 2.4 |

Two capacitors of $3.0\ \mu\text{F}$ and $6.0\ \mu\text{F}$ are connected in series across a $12\ \text{V}$ battery.

Find the equivalent capacitance and the charge on each capacitor.

Solution

For capacitors in series,

\frac{1}{C_{\text{eq}}} = \frac{1}{3.0} + \frac{1}{6.0} = \frac{1}{2.0}

C_{\text{eq}} = 2.0\ \mu\text{F}

The charge stored by the series combination is

Q = C_{\text{eq}}V = (2.0\ \mu\text{F})(12\ \text{V}) = 24\ \mu\text{C}

In series, each capacitor has the same charge, so each one stores $24\ \mu\text{C}$.

2.5Kirchhoff Loop with Power

Exam II | Problem 2.5 |

Kirchhoff's rules, Power

A $12\ \text{V}$ battery is connected in series with a $2\ \Omega$ resistor and a $4\ \Omega$ resistor.

Find the circuit current and the power dissipated by the $4\ \Omega$ resistor.

Solution

The total resistance is

R_{\text{tot}} = 2 + 4 = 6\ \Omega

So the current is

I = \frac{V}{R_{\text{tot}}} = \frac{12}{6} = 2\ \text{A}

The power in the $4\ \Omega$ resistor is

P = I^2R = (2)^2(4) = 16\ \text{W}

2.6RC Charging After One Time Constant

Exam II | Problem 2.6 |

An RC circuit has $R = 2.0\ \text{k}\Omega$, $C = 100\ \mu\text{F}$, and is connected to a $12\ \text{V}$ battery.

Find the time constant and the charge on the capacitor after one time constant has passed.

Solution

The time constant is

\tau = RC = (2.0\times 10^3)(100\times 10^{-6}) = 0.20\ \text{s}

The final charge is

Q_{\infty} = CV = (100\times 10^{-6})(12) = 1.2\times 10^{-3}\ \text{C}

After one time constant, the charge is about $63.2\%$ of the final value:

Q(\tau) = 0.632Q_{\infty}

Q(\tau) \approx 0.632(1.2\times 10^{-3}) = 7.6\times 10^{-4}\ \text{C}

2.7Magnetic Field of a Long Straight Wire

Exam II | Problem 2.7 |

Ampere's law, Magnetic field

A long straight wire carries a current of $8.0\ \text{A}$.

What is the magnetic field magnitude $5.0\ \text{cm}$ from the wire?

Solution

For a long straight wire,

B = \frac{\mu_0 I}{2\pi r}

Substitute:

B = \frac{(4\pi\times 10^{-7})(8.0)}{2\pi(0.050)}

B = 3.2\times 10^{-5}\ \text{T}

2.8Motion in a Uniform Magnetic Field

Exam II | Problem 2.8 |

A proton moves perpendicular to a uniform magnetic field of magnitude $0.50\ \text{T}$ with speed $2.0\times 10^5\ \text{m/s}$.

What is the radius of its circular path?

Solution

For circular motion in a magnetic field,

r = \frac{mv}{|q|B}

Using the proton mass $m = 1.67\times 10^{-27}\ \text{kg}$ and charge $q = 1.60\times 10^{-19}\ \text{C}$:

r = \frac{(1.67\times 10^{-27})(2.0\times 10^5)}{(1.60\times 10^{-19})(0.50)}

r \approx 4.2\times 10^{-3}\ \text{m}

Difficulty

Applied

3.1Field from an Infinite Line Charge

Final | Problem 3.1 |

Gauss's law, Electric field

An infinite line of charge has linear charge density

\lambda = 5.0\ \text{nC/m}.

What is the electric field magnitude $0.20\ \text{m}$ from the line?

Solution

For an infinite line charge,

E = \frac{\lambda}{2\pi\varepsilon_0 r}

Substitute:

E = \frac{5.0\times 10^{-9}}{2\pi(8.854\times 10^{-12})(0.20)}

E \approx 4.5\times 10^2\ \text{N/C}

Since the line charge is positive, the field points away from the line.

3.2Magnetic Field Inside a Solenoid

Final | Problem 3.2 |

Ampere's law, Magnetic field

A solenoid has $900$ turns over a length of $0.30\ \text{m}$ and carries a current of $2.0\ \text{A}$.

What is the magnetic field inside the solenoid?

Solution

First find the number of turns per unit length:

n = \frac{900}{0.30} = 3000\ \text{m}^{-1}

For a long solenoid,

B \approx \mu_0 n I

Substitute:

B = (4\pi\times 10^{-7})(3000)(2.0)

B \approx 7.5\times 10^{-3}\ \text{T}

3.3Direction of an Induced Current

Final | Problem 3.3 |

Faraday's law, Lenz's law

A circular loop lies in the page. The magnetic field through the loop points into the page and is increasing.

What is the direction of the induced current?

Solution

The increasing flux into the page must be opposed by an induced field out of the page.

Using the right-hand rule, a field out of the page is produced by a counterclockwise current.

So the induced current is counterclockwise.

3.4Motional EMF in a Sliding Bar

Final | Problem 3.4 |

A metal bar of length $0.40\ \text{m}$ slides at $5.0\ \text{m/s}$ through a magnetic field of magnitude $0.80\ \text{T}$. The bar and its motion are perpendicular to the field. The circuit resistance is $2.0\ \Omega$.

Find the induced emf and the current in the circuit.

Solution

The motional emf is

\mathcal{E} = BLv

\mathcal{E} = (0.80)(0.40)(5.0) = 1.6\ \text{V}

Then use Ohm's law:

I = \frac{\mathcal{E}}{R} = \frac{1.6}{2.0} = 0.80\ \text{A}

3.5Current Growth in an RL Circuit

Final | Problem 3.5 |

An RL circuit has resistance $4.0\ \Omega$, inductance $0.20\ \text{H}$, and final current $3.0\ \text{A}$.

What is the time constant, and what is the current after $0.05\ \text{s}$?

Solution

The time constant is

\tau = \frac{L}{R} = \frac{0.20}{4.0} = 0.050\ \text{s}

For current growth,

I(t) = I_{\infty}\left(1-e^{-tR/L}\right)

At $t = 0.05\ \text{s}$,

I = 3.0\left(1-e^{-1}\right)

I \approx 1.9\ \text{A}

Difficulty

Challenge

4.1Charging Capacitor and the Displacement Current

Final | Problem 4.1 |

Maxwell's equations, Current circuits

A capacitor is charging in a circuit, and the conduction current in the wire is $0.40\ \text{A}$.

According to Maxwell's equations, what must the displacement current term between the capacitor plates be, and why?

Solution

The Ampere-Maxwell law is

\oint \mathbf{B}\cdot d\mathbf{\ell} = \mu_0 I_{\text{enc}} + \mu_0\varepsilon_0 \frac{d\Phi_E}{dt}

For the magnetic field prediction to be the same no matter which surface is chosen, the displacement current term must match the conduction current.

So the displacement current is

0.40\ \text{A}

in magnitude.

4.2Zero Potential, Nonzero Field

Final | Problem 4.2 |

Two charges, $+2.0\ \text{nC}$ and $-2.0\ \text{nC}$, are placed on a line, each $0.10\ \text{m}$ from the midpoint.

What are the electric potential and the electric field at the midpoint?

Solution

The electric potential adds as a scalar:

V = k\frac{2.0\times 10^{-9}}{0.10} + k\frac{-2.0\times 10^{-9}}{0.10} = 0

So the potential at the midpoint is zero.

For the field, each charge contributes a field of magnitude

E_1 = 9.0\times 10^9 \frac{2.0\times 10^{-9}}{(0.10)^2} = 1800\ \text{N/C}

At the midpoint, both fields point from the positive charge toward the negative charge, so they add:

E = 3600\ \text{N/C}

The field points toward the negative charge.

4.3Electric and Magnetic Flux Through a Closed Surface

Final | Problem 4.3 |

Gauss's law, Maxwell's equations

A closed surface encloses a net charge of $+3.0\ \text{nC}$ and also surrounds a bar magnet.

What are the net electric flux and the net magnetic flux through the surface?

Solution

By Gauss's law for electricity,

\Phi_E = \frac{Q_{\text{enc}}}{\varepsilon_0}

\Phi_E = \frac{3.0\times 10^{-9}}{8.854\times 10^{-12}} \approx 3.4\times 10^2\ \text{N}\cdot\text{m}^2/\text{C}

By Gauss's law for magnetism,

\Phi_B = 0

So the surface has nonzero electric flux, but zero net magnetic flux.

4.4Faraday Sign Convention

Final | Problem 4.4 |

Faraday's law, Lenz's law

A circular loop lies in the page. The magnetic field through the loop points out of the page and is increasing.

Take counterclockwise circulation as the positive loop direction.

What is the sign of the induced emf, and what is the actual current direction?

Solution

With counterclockwise chosen as positive, the associated area vector points out of the page.

The magnetic flux is therefore positive, and because the field is increasing, $d\Phi_B/dt > 0$.

Faraday's law gives

\mathcal{E} = -\frac{d\Phi_B}{dt}

So the induced emf is negative.

A negative emf means the actual current goes opposite the positive loop direction, so the current is clockwise.