Physics 9702/41 — May/June 2023

Background Concept

A field is a way of describing how one object influences the space around it.

For gravity, we describe the effect of masses using the gravitational field strength $g$. It is defined so that if a small test mass $m$ is placed at a point in the field and experiences a force $F$, then

$$g = \frac{F}{m}$$

The test mass must be small so it does not significantly change the field.

Understanding the Question

You are asked to define gravitational field, so the mark is for a precise statement, typically in terms of “force per unit mass”.

Approach

Give the standard definition of field strength: force divided by the test quantity (mass for gravitational fields).

Step-by-Step Reasoning

• Place a small test mass $m$ at the point.
• It experiences a gravitational force $F$.
• Define the field strength as $g = F/m$.
• State this definition in words.

Key Takeaways

• Gravitational field strength is defined by force per unit mass.
• Using a “small test mass” avoids altering the field.

Common Mistakes

• Defining it as “force on an object” without the “per unit mass”.
• Confusing gravitational field strength $g$ with gravitational force $F$.

Things to Be Careful About

• The phrase “per unit mass” (or an equivalent equation $g = F/m$) is essential for full credit.
• Ensure the definition applies at a point in the field (not just generally around a planet).

Background Concept

Electric fields describe the influence of charges on the space around them.

The electric field strength $E$ is defined so that if a small positive test charge $q$ is placed at a point and experiences a force $F$, then

$$E = \frac{F}{q}$$

The direction of $E$ is defined as the direction of the force on a positive test charge.

Understanding the Question

This is a 1-mark “define” question. You must mention “force per unit positive charge” (or give $E = F/q$ with words identifying a positive test charge).

Approach

State the standard definition of electric field strength at a point.

Step-by-Step Reasoning

• Consider a small positive test charge $q$ at the point.
• It experiences an electric force $F$.
• Define $E$ by $E = F/q$.
• Convert to a clear sentence definition.

Key Takeaways

• Electric field strength is force per unit positive charge.
• The “positive test charge” convention sets the direction of the field.

Common Mistakes

• Missing “positive” when defining the direction.
• Writing “force per coulomb” without saying it is on a test charge at a point.

Things to Be Careful About

• Do not confuse $E$ (field strength) with $V$ (electric potential).
• State “at a point” or equivalent, to show it is a local definition.

Background Concept

Potential is energy per unit test quantity:

• Gravitational potential $\phi$ is gravitational potential energy per unit mass.
• Electric potential $V$ is electric potential energy per unit charge.

For a point mass $M$, gravitational potential at distance $r$ (taking zero at infinity) is

$$\phi = -\frac{GM}{r}$$

For a point charge $Q$, electric potential at distance $r$ (taking zero at infinity) is

$$V = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r}$$

Both share the same distance dependence ($\propto 1/r$), but they differ in sign behaviour because gravity is always attractive while electric forces can be attractive or repulsive.

Understanding the Question

You must state:

• one similarity between gravitational potential (point mass) and electric potential (point charge), and
• one difference.

A good similarity is the $1/r$ dependence. A good difference is the sign: gravity gives a negative potential (with zero at infinity), but electric potential depends on whether $Q$ is positive or negative.

Approach

Recall the standard expressions for potentials of point sources and compare:

• the factor in front (constants),
• the dependence on $r$,
• the sign conventions.

Step-by-Step Reasoning

• Write forms: $\phi = -GM/r$ and $V = (1/4\pi\varepsilon_0)\,Q/r$.
• Similarity: both are proportional to $1/r$ and approach $0$ as $r \to \infty$.
• Difference: $\phi$ is negative for a positive mass (because work must be done to separate masses to infinity), but $V$ has the same sign as $Q$ (positive for $+Q$, negative for $-Q$).

Key Takeaways

• Potentials for point sources have $1/r$ dependence.
• Electric potential can be positive or negative; gravitational potential (with the usual reference at infinity) is always negative.

Common Mistakes

• Saying “both are always negative”: only gravitational potential is always negative (with the usual convention).
• Giving a similarity like “both are fields”: potentials are not fields; they are scalar potentials.

Things to Be Careful About

• Make sure you are comparing potential (a scalar) not field strength.
• State only one similarity and one difference clearly, as requested, rather than multiple vague statements.

Background Concept

For spherically symmetric sources, the external field behaves as if all the source were concentrated at the centre.

• Gravitational field strength due to a mass $M$ at distance $r$:

$$g = \frac{GM}{r^2}$$

• Electric field strength due to a charge $Q$ at distance $r$:

$$E = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}$$

At the surface of a sphere, $r = R$ (the radius).

Understanding the Question

You are told an isolated uniform conducting sphere has mass $M$ and charge $Q$. At its surface the gravitational field is $g$ and the electric field is $E$.

You must show that $M/Q$ is proportional to $g/E$, i.e.

$$\frac{M}{Q} = \alpha \frac{g}{E}$$

and identify that $\alpha$ is a constant (independent of $M$, $Q$, $R$).

Approach

Write $g$ and $E$ at the surface in terms of $M$, $Q$ and $R$. Because both have the same $1/R^2$ dependence, taking a ratio will cancel $R^2$. Then rearrange into the requested form and read off the constant $\alpha$.

Step-by-Step Reasoning

1. For the sphere’s surface ($r=R$), use the inverse-square expressions:

$$g = \frac{GM}{R^2}$$ $$E = \frac{1}{4\pi\varepsilon_0}\frac{Q}{R^2}$$

1. Take the ratio $g/E$ to eliminate $R^2$:

$$\frac{g}{E} = \frac{GM/R^2}{\left(\frac{1}{4\pi\varepsilon_0}\right)Q/R^2}$$

The $R^2$ cancels, giving

$$\frac{g}{E} = 4\pi\varepsilon_0 G\frac{M}{Q}$$

1. Rearrange to make $M/Q$ the subject:

$$\frac{M}{Q} = \frac{1}{4\pi\varepsilon_0 G}\frac{g}{E}$$

So the constant is

$$\alpha = \frac{1}{4\pi\varepsilon_0 G}$$

Key Takeaways

• Both gravitational and electric fields outside a spherical source follow an inverse-square law.
• Ratios are a powerful way to remove a common unknown (here, the radius $R$).
• The proportionality constant here depends only on universal constants $G$ and $\varepsilon_0$.

Common Mistakes

• Using $E = V/d$ (only for uniform fields) instead of the point/spherical field expression.
• Forgetting the factor $1/(4\pi\varepsilon_0)$ in the electric field.
• Not cancelling $R^2$ correctly.

Things to Be Careful About

• $E$ here is the field at the surface, so you must use $r=R$.
• Keep the algebra symbolic until the final expression so it is clear that $\alpha$ is a constant.

Background Concept

Once you have expressed a constant symbolically, you can find its numerical value by substituting accepted values of physical constants.

Here,

$$\alpha = \frac{1}{4\pi\varepsilon_0 G}$$

The units should come out as $\text{kg}^2\ \text{C}^{-2}$ because $4\pi\varepsilon_0 G$ has units $\text{C}^2\ \text{kg}^{-2}$.

Understanding the Question

You are asked to show the numerical value of $\alpha$. You use the expression from (b)(i) and substitute $\varepsilon_0$ and $G$.

Approach

• Write $\alpha$ in terms of $\varepsilon_0$ and $G$.
• Substitute the standard values.
• Calculate carefully and present in standard form with units.

Step-by-Step Reasoning

Start with

$$\alpha = \frac{1}{4\pi\varepsilon_0 G}$$

Substitute:

$$\alpha = \frac{1}{4\pi (8.85 \times 10^{-12})(6.67 \times 10^{-11})}$$

Calculate the product in the denominator:

• numerical part: $4\pi \times 8.85 \times 6.67 \approx 741$
• powers of ten: $10^{-12} \times 10^{-11} = 10^{-23}$

So

$$4\pi\varepsilon_0 G \approx 741 \times 10^{-23} = 7.41 \times 10^{-21}$$

Hence

$$\alpha \approx \frac{1}{7.41 \times 10^{-21}} \approx 1.35 \times 10^{20}$$

Units: the result is

$$\alpha \approx 1.35 \times 10^{20}\ \text{kg}^2\ \text{C}^{-2}$$

Key Takeaways

• Use the derived symbolic expression first, then substitute constants.
• Keep track of powers of ten separately to avoid calculator mistakes.
• Always attach units to constants when asked.

Common Mistakes

• Missing $4\pi$ in the expression.
• Using an incorrect value for $\varepsilon_0$ or $G$.
• Writing the unit incorrectly (e.g. $\text{kg C}^{-2}$ instead of $\text{kg}^2\ \text{C}^{-2}$).

Things to Be Careful About

• Standard form: $1.35 \times 10^{20}$, not a long decimal.
• Do not round too early; keep at least 3 s.f. in intermediate steps to obtain the stated value.

Background Concept

From part (b), for any conducting sphere where the gravitational field at the surface is $g$ and the electric field at the surface is $E$,

$$\frac{M}{Q} = \alpha \frac{g}{E}$$

This links the ratio $M/Q$ of the sphere to the ratio $g/E$ at its surface.

Rearranging is often the key skill: you need to make the required quantity (here $E$) the subject.

Electric field strength has units $\text{N C}^{-1}$ (equivalently $\text{V m}^{-1}$).

Understanding the Question

You are told:

• Earth mass $M = 5.98 \times 10^{24}\ \text{kg}$,
• Earth charge $Q = -4.80 \times 10^{5}\ \text{C}$,
• treat Earth as a uniform conducting sphere.

Using the relationship from (b) and the known surface gravitational field strength ($g \approx 9.81\ \text{N kg}^{-1}$), find $E$ at the surface and state a unit.

Approach

Rearrange

$$\frac{M}{Q} = \alpha \frac{g}{E}$$

to get $E$ in terms of $\alpha$, $g$, $Q$, and $M$, then substitute values. Because the question asks for field strength, give the magnitude (a positive number) with unit; the sign is handled in the next part (direction).

Step-by-Step Reasoning

Start with

$$\frac{M}{Q} = \alpha \frac{g}{E}$$

Multiply both sides by $E$:

$$E\frac{M}{Q} = \alpha g$$

So

$$E = \alpha g \frac{Q}{M}$$

Substitute the given values:

$$E = (1.35 \times 10^{20})(9.81)\frac{-4.80 \times 10^{5}}{5.98 \times 10^{24}}$$

Compute the magnitude:

• Numerator scale: $1.35 \times 9.81 \times 4.80 \approx 63.6$
• Powers: $10^{20} \times 10^{5} / 10^{24} = 10^{1}$

So the magnitude is approximately

$$|E| \approx \frac{63.6 \times 10^{1}}{5.98} \approx 1.06 \times 10^{2}$$

Hence

$$|E| \approx 1.1 \times 10^{2}\ \text{N C}^{-1}$$

The negative sign in the raw calculation indicates direction (towards the negative Earth charge), which is addressed in part (c)(ii).

Key Takeaways

• Rearranging a proportionality is essential before substituting numbers.
• Electric field strength units: $\text{N C}^{-1}$ (or $\text{V m}^{-1}$).
• A negative value from $E = \alpha g Q/M$ indicates direction, not a “negative strength”.

Common Mistakes

• Using $Q/M$ instead of $M/Q$ (or not rearranging correctly).
• Forgetting to include the unit.
• Treating the negative sign as meaning “field strength is negative” rather than “field direction is opposite to an outward radial direction”.

Things to Be Careful About

• Use appropriate significant figures (typically 2 s.f. here, matching given data).
• Be consistent with the value of $g$ at Earth’s surface (commonly $9.81\ \text{N kg}^{-1}$).
• If you quote a negative $E$, you must also state the direction clearly; otherwise, give the magnitude and handle direction separately.

Background Concept

Field direction is defined by the direction of force on a positive test quantity:

• Gravitational field direction: direction of force on a test mass (mass is always positive), so it points towards the attracting mass.
• Electric field direction: direction of force on a positive test charge.

A negative source charge produces electric field lines that point towards the charge.

Understanding the Question

Earth’s surface charge is given as negative ($-4.80 \times 10^{5}\ \text{C}$). You must compare:

• direction of electric field at Earth’s surface, and
• direction of gravitational field at Earth’s surface.

Approach

Decide each direction separately using the sign/convention, then compare.

Step-by-Step Reasoning

• Gravitational field: Earth attracts masses, so the gravitational field at the surface points towards Earth’s centre (radially inward).
• Electric field: direction is the force on a positive test charge. A positive test charge is attracted to a negative Earth, so the electric field points towards the Earth (radially inward).

Therefore, both directions are the same.

Key Takeaways

• Electric field direction depends on sign of charge; gravitational field direction does not (mass is always attractive).
• Negative charge gives an inward electric field.

Common Mistakes

• Saying the electric field points outward because “it is a field coming from Earth” (field lines for negative charges end on the charge).
• Confusing field direction with the sign in a calculation without stating what the sign means.

Things to Be Careful About

• The electric field direction is defined using a positive test charge.
• At the surface, “towards Earth” means radially inward (towards the centre).