Physics 9702/43 — May/June 2011

Background Concept

A field is a way of describing forces that act at a distance (without direct contact). At each point in space, the field tells you what force would act on a suitable small test object placed there.

For example:

• gravitational field strength $g$ is defined by $g = F/m$ (force per unit mass),
• electric field strength $E$ is defined by $E = F/q$ (force per unit positive charge).

Understanding the Question

The question asks for a general meaning of a field of force, not specifically gravitational or electric. So you should define it as a region of space where a test object feels a force.

Approach

Give a short definition with the key idea:

• region of space,
• test object experiences a force.

Step-by-Step Reasoning

1. Identify that a “field of force” is not the force itself but the space around a source (mass/charge) where effects are felt.
2. State the operational meaning: if you put a suitable test object in that region, it will experience a force.

Key Takeaways

• A field of force describes forces acting at a distance.
• The definition must mention a region of space and a force on a test object.

Common Mistakes

• Saying “a force acting at a distance” without mentioning a region of space.
• Confusing the field with field lines (field lines are just a representation).

Things to Be Careful About

• Keep it general: do not define only gravitational field or only electric field unless the question asks for that.
• A good definition is independent of the particular type of field.

Background Concept

Gravitational and electric interactions are both examples of inverse-square forces for point sources:

• Gravity (Newton’s law):

$$F_g = \frac{G m_1 m_2}{r^2}$$

• Electrostatic force (Coulomb’s law):

$$F_e = \frac{1}{4\pi\varepsilon_0}\frac{Q_1 Q_2}{r^2}$$

Both can be described by a field: a mass or charge creates a field around it, and another mass/charge placed in that field experiences a force.

Understanding the Question

You must give:

• one similarity between gravitational and electric fields,
• one difference between them.

Each statement should be clear and unambiguous (so the examiner can award marks easily).

Approach

Pick high-value comparisons that are always true for the basic models used at A Level:

• Similarity: both are long-range inverse-square radial fields for point sources.
• Difference: direction/type of force (always attractive for gravity; attraction/repulsion for electric).

Step-by-Step Reasoning

1. For similarity, look at the equations: both contain $1/r^2$, meaning the strength falls with the square of distance from a point source. This also implies a radial symmetry about the source.
2. For difference, compare the sign behaviour:
   • masses are always positive, so $F_g$ always produces attraction.
   • charges can be positive or negative, so $Q_1Q_2$ can be positive (repulsion) or negative (attraction).

Key Takeaways

• Both gravitational and electric fields obey inverse-square behaviour for point sources.
• Electric interaction can attract or repel; gravity (for normal matter) only attracts.

Common Mistakes

• Giving a vague similarity like “both have field lines” without stating a physical property.
• Saying “electric is stronger” as the only difference (it is true for protons, but strength depends on what you compare).
• Mixing up “field” and “force” wording (a field is the region/property; the force acts on a test object).

Things to Be Careful About

• Only one similarity and one difference are required, but each must be stated precisely.
• If you choose “inverse-square” as the similarity, ensure you specify it is for a point source / spherically symmetric situation.

Background Concept

For two point objects separated by distance $R$, both gravity and electrostatic interactions follow inverse-square laws.

• Coulomb’s law for two charges $Q_1$ and $Q_2$:

$$F_e = \frac{1}{4\pi\varepsilon_0}\frac{Q_1Q_2}{R^2}$$

• Newton’s law of gravitation for two masses $m_1$ and $m_2$:

$$F_g = \frac{Gm_1m_2}{R^2}$$

Here each proton has charge $+e$ and mass $m_p$.

Understanding the Question

Two protons are separated by $R$. You are asked for the ratio:

$$\frac{\text{electric force}}{\text{gravitational force}}$$

Because both forces vary as $1/R^2$, the ratio should not depend on $R$ (a useful check).

Approach

1. Write expressions for $F_e$ and $F_g$ for two protons.
2. Form $F_e/F_g$ and cancel the common factor $1/R^2$.
3. Substitute constants ($e$, $m_p$, $G$, and $1/4\pi\varepsilon_0$) and calculate, giving the answer in standard form.

Step-by-Step Reasoning

1. Electric force between two protons:

$$F_e = \frac{1}{4\pi\varepsilon_0}\frac{e\cdot e}{R^2} = \frac{1}{4\pi\varepsilon_0}\frac{e^2}{R^2}$$

1. Gravitational force between two protons:

$$F_g = \frac{Gm_p m_p}{R^2} = \frac{Gm_p^2}{R^2}$$

1. Form the ratio:

$$\frac{F_e}{F_g} = \frac{\left(\frac{1}{4\pi\varepsilon_0}\right)\frac{e^2}{R^2}}{\frac{Gm_p^2}{R^2}} = \frac{\left(\frac{1}{4\pi\varepsilon_0}\right)e^2}{Gm_p^2}$$

The $R^2$ cancels, confirming the ratio is independent of separation.

1. Substitute values:

• $\frac{1}{4\pi\varepsilon_0} = 8.99 \times 10^9\ \text{N m}^2\text{C}^{-2}$
• $e = 1.60 \times 10^{-19}\ \text{C}$
• $m_p = 1.67 \times 10^{-27}\ \text{kg}$
• $G = 6.67 \times 10^{-11}\ \text{N m}^2\text{kg}^{-2}$

Compute the squared terms:

$$e^2 = (1.60\times 10^{-19})^2 = 2.56\times 10^{-38}$$ $$m_p^2 = (1.67\times 10^{-27})^2 \approx 2.79\times 10^{-54}$$

Now evaluate:

$$\text{numerator} = (8.99\times 10^9)(2.56\times 10^{-38}) \approx 2.30\times 10^{-28}$$ $$\text{denominator} = (6.67\times 10^{-11})(2.79\times 10^{-54}) \approx 1.86\times 10^{-64}$$

So

$$\frac{F_e}{F_g} \approx \frac{2.30\times 10^{-28}}{1.86\times 10^{-64}} \approx 1.24\times 10^{36}$$

To 2 s.f., $1.2\times 10^{36}$.

Key Takeaways

• Both forces are inverse-square, so their ratio is independent of $R$.
• Comparing electric and gravitational forces between protons shows gravity is enormously weaker.

Common Mistakes

• Forgetting to square $e$ or $m_p$.
• Not cancelling $R^2$ and incorrectly trying to substitute a value for $R$.
• Using $\varepsilon_0$ instead of $1/(4\pi\varepsilon_0)$ (missing the $4\pi$ factor).
• Power-of-ten errors when dividing numbers in standard form.

Things to Be Careful About

• Keep constants in consistent standard form throughout.
• Use an appropriate number of significant figures (typically 2 or 3 for constants-based results).
• The ratio is dimensionless; if you end up with units, something has been handled incorrectly.