10th Class Physics Chapter 6 Notes PDF 2026 – Electrostatics

Definition

When two different materials are rubbed together, electrons transfer from one to the other — the material that holds its outer electrons less tightly loses them and becomes positively charged, while the material that attracts electrons more strongly gains them and becomes negatively charged. Which way electrons transfer depends on the two materials' relative tendency to hold onto electrons, not on some materials simply being 'chargeable' and others not.

Conductors (e.g. metals) allow charge to move freely through them, since they have free electrons; insulators (e.g. plastic, glass, rubber) do not, so charge stays wherever it's placed.

In a conductor placed in an electric field, free electrons redistribute until the field inside becomes zero. In an insulator, charges cannot move freely, so the material's own charge distribution stays largely fixed, though individual molecules may still align slightly (polarize).

Example 1 — a charged balloon and water

A balloon rubbed on dry hair is held near a thin stream of water, and the water bends toward it. What does this show?

The balloon becomes charged by friction (gains electrons from hair).
Water molecules are polar — each has a slightly positive and slightly negative end.
The charged balloon attracts the oppositely-charged end of nearby water molecules, pulling the stream toward it.

Answer: Water molecules are polar, so they respond to a nearby charge even without being charged themselves

Remember

A charged comb attracts paper pieces at first (by inducing an opposite charge on the near side of the paper), but the pieces fall off after a short time because some charge slowly transfers between the comb and the paper on contact, neutralizing the attraction.

Definition

Electrical breakdown occurs when the electric field in an insulator (like air) becomes strong enough to rip electrons free from its atoms, suddenly turning the insulator into a conductor for a brief moment — a visible spark or lightning bolt is this rapid discharge of built-up charge through the air.

Remember

Static electricity is dangerous around flammable dust or vapours (e.g. in grain silos or fuel-handling areas) because even a small spark from electrical breakdown can ignite them — a major real-world hazard behind dust explosions in factories, minimized by grounding equipment so charge cannot build up.

Rule

Coulomb's law: the electrostatic force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them — like charges repel, unlike charges attract, and this force gets weaker very quickly (as $1/ r^{2}$ ) as the charges move apart.

$F = \frac{k q _{1} q _{2}}{r ^{2}}$ , where $k \approx 9 \times 1 0^{9} N \cdot m^{2} / C^{2}$

Example 2 — Coulomb's force between two charges

Two point charges $q_{1} = 8 μ C$ and $q_{2} = 4 μ C$ are $120$ cm apart. Find the Coulomb force between them and state its nature.

$r = 1.2$ m, $q_{1} = 8 \times 1 0^{- 6}$ C, $q_{2} = 4 \times 1 0^{- 6}$ C.
$F = \frac{k q _{1} q _{2}}{r ^{2}} = \frac{( 9 \times 1 0 ^{9} ) ( 8 \times 1 0 ^{- 6} ) ( 4 \times 1 0 ^{- 6} )}{( 1.2 ) ^{2}}$ .
Numerator $= 9 \times 1 0^{9} \times 32 \times 1 0^{- 12} = 0.288$ .
$F = \frac{0.288}{1.44}$ .

Answer: $F = 0.2$ N, repulsive (both charges are positive, like charges repel)

Example 3 — force at a different separation

Two identical charges repel with force $0.2$ N when $9$ cm apart. Find the force when they are $3$ cm apart.

$F \propto \frac{1}{r ^{2}}$ , so $\frac{F _{2}}{F _{1}} = (\frac{r _{1}}{r _{2}})^{2}$ .
$\frac{F _{2}}{0.2} = (\frac{9}{3})^{2} = 9$ .
$F_{2} = 0.2 \times 9$ .

Answer: $F_{2} = 1.8$ N

Example 4 — repulsive force between equal spheres

Two equally charged spheres, each $3 μ C$ , are $2.0$ cm apart. Find the repulsive force each exerts on the other.

$r = 0.02$ m, $q_{1} = q_{2} = 3 \times 1 0^{- 6}$ C.
$F = \frac{( 9 \times 1 0 ^{9} ) ( 3 \times 1 0 ^{- 6} ) ^{2}}{( 0.02 ) ^{2}} = \frac{( 9 \times 1 0 ^{9} ) ( 9 \times 1 0 ^{- 12} )}{0.0004}$ .
$F = \frac{0.081}{0.0004}$ .

Answer: $F = 202.5$ N

Example 5 — finding charge from force

Two equal charges repel with $0.5$ N when $3$ cm apart. Find the charge on each sphere.

$F = \frac{k q ^{2}}{r ^{2}}$ , so $q^{2} = \frac{F r ^{2}}{k} = \frac{( 0.5 ) ( 0.03 ) ^{2}}{9 \times 1 0 ^{9}}$ .
$q^{2} = \frac{0.00045}{9 \times 1 0 ^{9}} = 5 \times 1 0^{- 14}$ .
$q = 5 \times 1 0^{- 14}$ .

Answer: $q \approx 2.24 \times 1 0^{- 7}$ C

Example 6 — finding the second charge

A force of $85$ N acts between two charges $3.2$ cm apart. One charge is $25 μ C$ . Find the other charge.

$r = 0.032$ m, $q_{1} = 25 \times 1 0^{- 6}$ C.
$q_{2} = \frac{F r ^{2}}{k q _{1}} = \frac{( 85 ) ( 0.032 ) ^{2}}{( 9 \times 1 0 ^{9} ) ( 25 \times 1 0 ^{- 6} )}$ .
$q_{2} = \frac{0.087}{225 , 000}$ .

Answer: $q_{2} \approx 3.87 \times 1 0^{- 7}$ C

Definition

Electric field intensity at a point is the force per unit positive charge placed there — a vector, pointing in the direction a positive test charge would be pushed. Field lines point away from a positive charge and into a negative charge; for a parallel-plate capacitor, they run straight and evenly spaced from the positive plate to the negative plate, showing the field is uniform between the plates.

$E = \frac{F}{q}$ (from a test charge), or $E = \frac{k q}{r ^{2}}$ (field of a point charge $q$ at distance $r$ )

Example 7 — field from force on a test charge

A test charge $25 μ C$ placed in a field feels a force of $0.500$ N. Find the field strength.

$E = \frac{F}{q} = \frac{0.500}{25 \times 1 0 ^{- 6}}$ .

Answer: $E = 20, 000$ N/C

Example 8 — field of a point charge

Find the electric field strength at $0.5$ m from a point charge $q = 10 μ C$ in air.

$E = \frac{k q}{r ^{2}} = \frac{( 9 \times 1 0 ^{9} ) ( 10 \times 1 0 ^{- 6} )}{( 0.5 ) ^{2}} = \frac{90 , 000}{0.25}$ .

Answer: $E = 360, 000$ N/C

Definition

A gold-leaf electroscope detects and roughly measures charge: bringing a charged object near its cap induces charge redistribution down the metal rod, causing the two thin gold leaves at the bottom to repel and spread apart — the more they diverge, the greater the magnitude of charge. Touching the cap with a known charge (or comparing how the leaves respond to a charge of known sign) can also reveal the sign of an unknown charge already on the electroscope.

Quick Recap

●Friction transfers electrons; conductors let charge move freely, insulators don't
●Electrical breakdown: field strong enough to rip electrons free $\to$ spark/lightning
●Coulomb's law: $F = \frac{k q _{1} q _{2}}{r ^{2}}$ , $k \approx 9 \times 1 0^{9} N \cdot m^{2} / C^{2}$ ; like repels, unlike attracts
●Electric field: $E = F / q = k q / r^{2}$ ; field lines point away from $+$ , into $-$
●Gold-leaf electroscope: leaf divergence shows the magnitude of charge

10th Class Physics Chapter 6 Notes

Charging by Friction, Conductors, and Insulators

Electrical Breakdown

Coulomb's Law

Electric Field

The Gold-Leaf Electroscope

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