Chapter 15: Electrostatics (Numerical Problems)

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Chapter Overview & SLOs

How is the magnitude of an electric field determined? The electric field strength (E) at a point is calculated by dividing the electrostatic force (F) acting on a test charge by the magnitude of that charge (q), expressed as E = F/q.

What is the unit of electric field strength? The unit for electric field strength is Newtons per Coulomb (N/C).

How does the sign of a charge affect the direction of the force? A positive charge experiences a force in the same direction as the electric field, while a negative charge (such as an electron) experiences a force in the opposite direction.

What are the key formulas for electrostatics numericals?

| Formula | Description | |---------|-------------| | E = F/q | Electric field strength = Force / Test charge | | F = qE | Force on a charge in an electric field | | q = F/E | Finding unknown test charge |

Unit conversions: 1 μC (micro-Coulomb) = 1 × 10⁻⁶ C. Always convert μC to C before calculations.

Example 1 - Electric field strength: For a 30 μC test charge experiencing a force of 0.600 N: q = 30 × 10⁻⁶ C. E = F/q = 0.600 / (30 × 10⁻⁶) = 2.00 × 10⁴ N/C.

Example 2 - Force on negative charge: A charge of -1.0 × 10⁻⁶ C is in a field of 1.7 × 10⁶ N/C directed to the right. Magnitude of force F = |q|E = (1.0 × 10⁻⁶)(1.7 × 10⁶) = 1.7 N. Since the charge is negative, the force is directed opposite to the field (to the left).

Example 3 - Doubling the field: If the field strength is doubled to 3.4 × 10⁶ N/C, for a positive charge the force is F = qE = (1.0 × 10⁻⁶)(3.4 × 10⁶) = 3.4 N directed to the right (same direction as field).

Example 4 - Finding unknown test charge: A charge experiences 0.5 N in a 2.0 N/C field. q = F/E = 0.5/2.0 = 0.25 C.

Example 5 - Force on an electron: An electron has charge q = -1.6 × 10⁻¹⁹ C. In a uniform field of 20 N/C, the force magnitude is F = |q|E = (1.6 × 10⁻¹⁹) × 20 = 3.2 × 10⁻¹⁸ N. The direction is opposite to the field because the electron is negatively charged.

Important note for force vectors: Always specify the direction (left/right, up/down, with/against the field) when stating the force on a charge. The magnitude is positive, but direction is essential.

These notes are strictly aligned with the Student Learning Outcomes (SLOs) for the FBISE 2026 annual examination.

• How do we calculate the electric field strength from a given force and charge? For a 30 μC test charge experiencing a force of 0.600 N, the electric field magnitude is $E = 0.600 / (30 \times 10^{-6}) = 2.00 \times 10^4\text{ N/C}$.
• How is the force on a charge determined when the field strength is modified? If a charge of $-1.0 \times 10^{-6}\text{ C}$ is in a field of $1.7 \times 10^6\text{ N/C}$ (right), it feels a force of 1.7 N to the left; if the field is doubled, the force on a similar positive charge becomes 3.4 N to the right.
• How do we find the magnitude of an unknown test charge? By rearranging the field formula to $q = F/E$, a charge experiencing 0.5 N in a 2.0 N/C field is calculated to be 0.25 C.
• How do we calculate the force acting specifically on an electron? Given an electron's charge ($-1.6 \times 10^{-19}\text{ C}$) in a uniform field of 20 N/C, the force magnitude is $F = (1.6 \times 10^{-19}) \times 20 = 3.2 \times 10^{-18}\text{ N}$, directed opposite to the field.

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