Electricity-And-Magnetism · Unit 2 · Video 5 · Interactive Practice

Gravitational, Electric and Magnetic Fields

IKey Formulas

Formula	Name	What it says
$\vec F_g = -G\,\dfrac{Mm}{r^2}\,\hat r$	Newton's law of gravitation	Always attractive, $\propto 1/r^2$ ; $G = 6.67 \times 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$
$\vec g = \lim\limits_{m \to 0} \dfrac{\vec F_g}{m} = -G\,\dfrac{M}{r^2}\,\hat r$	Gravitational field	Points toward $M$ ; at Earth's surface $g = GM/R_E^2 \approx 9.8\ \text{N/kg}$
$\vec F_e = k_e\,\dfrac{Qq}{r^2}\,\hat r$	Coulomb's law	$Qq > 0$ repels, $Qq < 0$ attracts; $k_e = 8.99 \times 10^{9}\ \text{N}\cdot\text{m}^2/\text{C}^2$
$\vec E = \lim\limits_{q \to 0} \dfrac{\vec F_e}{q} = k_e\,\dfrac{Q}{r^2}\,\hat r, \quad \vec F_e = q\vec E$	Electric field	Away from $Q > 0$ , toward $Q < 0$ ; unit $\text{N/C}$

Key Insight: A field is force per unit test object — divide out $m$ or $q$ and what remains belongs to the source alone. The magnetic field of a current circles the wire, in the sense given by the right-hand rule.

IIVisualization 1 — Divide Out the Test Object

The force grows with the test mass or charge; the quotient $\vec F/m$ or $\vec F/q$ depends only on the source and $r$ .

Source Test object q = 1 nC

IIIVisualization 2 — The Field Sets the Acceleration, Not the Velocity

In a uniform field $\vec g$ , a stone moves along the field only if it starts with velocity along $\vec g$ .

Launch speed v₀ 6 m/s Time 0.71 s

IVVisualization 3 — Which Way Does the Magnetic Field Circle?

A current, or a moving charge, is circled by $\vec B$ in the sense of the right-hand rule: thumb along $I$ , fingers along $\vec B$ .

Source

VQuiz Questions

Problem 1 · Gravitational Field Above the Earth

Given: At Earth's surface $g = GM/R_E^2 = 9.81\ \text{N/kg}$ . A satellite orbits at an altitude of one Earth radius, $R_E$ , above the surface — find the magnitude of the gravitational field there.

✅ Correct! The distance from Earth's centre doubles to

2R_E

, so the field drops to one quarter:

9.81/4 = 2.45\ \text{N/kg}

❌ Not quite. In

g = GM/r^2

r

is measured from Earth's centre, not from the surface: here

r = R_E + R_E = 2R_E

❌ Close, but… the field falls off as

1/r^2

, not

1/r

: doubling

r

divides

g

4

, not by

2

❌ Not quite. Find

r

from Earth's centre first (

r = 2R_E

), then scale

g

(R_E/r)^2

Show solution

The field magnitude at distance $r$ from Earth's centre is $g(r) = GM/r^2$ . The satellite is at altitude $R_E$ , so $r = R_E + R_E = 2R_E$ :

g(2R_E) = \frac{GM}{(2R_E)^2} = \frac{1}{4}\cdot\frac{GM}{R_E^2} = \frac{9.81\ \text{N/kg}}{4} = 2.45\ \text{N/kg}

The mass of the satellite never enters: the field belongs to the Earth alone.

Problem 2 · Field From a Force on a Negative Charge

Given: A test charge $q = -2.0\ \text{nC}$ placed at point $P$ feels a force of magnitude $3.6 \times 10^{-5}\ \text{N}$ pointing east — find the electric field at $P$ .

✅ Correct!

\vec E = \vec F/q

, and dividing by a negative

q

reverses the direction.

❌ Close, but check the direction. The field points the way a positive test charge would be pushed. Since

q < 0

\vec E = \vec F/q

points opposite to

\vec F

❌ Not quite. That is

Fq

, not

F/q

. The field is force per unit charge.

❌ Not quite. Convert

\text{nC}

to coulombs:

2.0\ \text{nC} = 2.0 \times 10^{-9}\ \text{C}

Show solution

Divide the force by the test charge, keeping its sign. Take east as $+\hat x$ :

\vec E = \frac{\vec F}{q} = \frac{(3.6 \times 10^{-5}\ \text{N})\,\hat x}{-2.0 \times 10^{-9}\ \text{C}} = -1.8 \times 10^{4}\ \text{N/C}\ \hat x

So $|\vec E| = 1.8 \times 10^{4}\ \text{N/C}$ , pointing west. A positive charge placed at $P$ would be pushed west; the negative test charge is pushed east, against the field.

Problem 3 · Field First, Then Force

Given: A point charge $Q = +3.0\ \text{nC}$ and a point $P$ at $r = 0.30\ \text{m}$ from it; $k_e = 8.99 \times 10^{9}\ \text{N}\cdot\text{m}^2/\text{C}^2$ — find the field at $P$ , then the force on a charge $q = -5.0\ \text{nC}$ placed at $P$ .

What is the field magnitude at P?

What is the force on q?

✅ Correct!

Q

makes the field (

300\ \text{N/C}

, outward); the field acts on

q

\vec F = q\vec E

, reversed because

q < 0

❌ Close, but…

89.9

k_eQ/r

. The field falls off as

1/r^2

: divide by

(0.30)^2 = 0.09

❌ Check the direction.

\vec E

points away from the positive

Q

, but

q

is negative, so

\vec F = q\vec E

points back toward

Q

: unlike charges attract.

❌ Check the field. Use

E = k_eQ/r^2

with

Q = 3.0 \times 10^{-9}\ \text{C}

and

r^2 = 0.09\ \text{m}^2

; the test charge does not enter.

❌ Check the force. Multiply:

|F| = |q|E = (5.0 \times 10^{-9}\ \text{C})(300\ \text{N/C})

Show solution

Step 1 — the field of $Q$ at $P$ :

E = k_e\frac{Q}{r^2} = \frac{(8.99 \times 10^{9})(3.0 \times 10^{-9})}{(0.30)^2} = \frac{26.97}{0.09} \approx 3.0 \times 10^{2}\ \text{N/C}

Since $Q > 0$ , $\vec E$ points radially away from $Q$ .

Step 2 — the force on $q$ :

\vec F = q\vec E, \qquad |F| = (5.0 \times 10^{-9}\ \text{C})(299.7\ \text{N/C}) \approx 1.5 \times 10^{-6}\ \text{N}

Because $q < 0$ , $\vec F$ is opposite to $\vec E$ : it points toward $Q$ . Check with Coulomb's law directly: $k_e|Qq|/r^2 = (8.99 \times 10^{9})(3.0 \times 10^{-9})(5.0 \times 10^{-9})/0.09 \approx 1.5 \times 10^{-6}\ \text{N}$ ✓, and $Qq < 0$ means attraction ✓.

Problem 4 · Magnetic Field of an Electron Beam

Given: A beam of electrons travels to the right across the page, at a speed small compared with $c$ — find the direction of the magnetic field at a point directly above the beam.

✅ Correct! Negative charges moving right are a current to the left; with the thumb pointing left, the fingers go into the page above the beam.

❌ Close, but… that is the field of a current to the right. Electrons are negative, so their current points opposite to their velocity, and the circling reverses.

❌ Not quite. The magnetic field of a moving charge circles its line of motion; it has no component along the beam.

❌ Not quite. A radial field is the electron's electric field. The magnetic field circles the line of motion.

Show solution

Current is the direction positive charge flows. Electrons ( $q < 0$ ) moving right are equivalent to a current $I$ pointing left.

Right-hand rule: thumb to the left along $I$ . Above the beam the fingers curl into the page; below the beam they come out of it.

Optional check, beyond the video (which uses only the right-hand rule): the field direction is $\hat I \times \hat r$ , with $\hat r$ pointing from the beam to the point: $\hat I \times \hat r = (-\hat x) \times (+\hat y) = -\hat z$ , which is into the page.

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