Gravitational Field, Electric Force, and Current Electricit

Gravitational Field, Electric Force, and Current Electricity

Gravitational Field

A gravitational field is a region where a mass experiences a force due to gravity. It influences planetary motion, tides, and free-falling objects.

Newton’s Law of Universal Gravitation

$$F = \frac{G M_1 M_2}{r^2}$$

where:

• $M_1, M_2$ = Masses of the objects (kg)
• $r$ = Distance between masses (m)
• $G$ = Gravitational constant ($6.67 \times 10^{-11} \, \text{N} \cdot \text{m}^2 \text{kg}^{-2}$)

---

Electric Force (Coulomb’s Law)

$$F = \frac{K q_1 q_2}{r^2}$$

where:

• $q_1, q_2$ = Charges (C)
• $r$ = Distance between charges (m)
• $K$ = Coulomb’s constant ($9.0 \times 10^9 \, \text{N} \cdot \text{m}^2 \text{C}^{-2}$)

---

Electromotive Force (E.M.F.)

Electromotive force ($E$) is the energy supplied per coulomb of charge by a power source.

$$E = V + Ir$$

where:

• $E$ = Electromotive force (V)
• $V$ = Terminal voltage (V)
• $I$ = Current (A)
• $r$ = Internal resistance of the source ($\Omega$)

Power Output of a Battery

$$P = EI$$

where:

• $P$ = Power (W)
• $I$ = Current (A)

---

Worked Examples

Gravitational Field – 5 Questions & Solutions

Example 1:

Question: The mass of the Earth is $5.98 \times 10^{24}$ kg, and its radius is $6.38 \times 10^6$ m. Calculate the acceleration due to gravity on the Earth's surface.

Solution:
Using $g = \frac{G M_e}{r_e^2}$,

$$g = \frac{(6.67 \times 10^{-11}) (5.98 \times 10^{24})}{(6.38 \times 10^6)^2}$$ $$g = 9.8 \, \text{m/s}^2$$

(Include 4 more similar problems with varying values.)

---

Electric Force – 5 Questions & Solutions

Example 1:

Question: Two charges, $5 \times 10^{-6} C$ and $3 \times 10^{-6} C$, are 0.2 m apart. Calculate the electric force between them.

Solution:
Using $F = \frac{K q_1 q_2}{r^2}$,

$$F = \frac{(9.0 \times 10^9) (5 \times 10^{-6}) (3 \times 10^{-6})}{(0.2)^2}$$ $$F = 3.375 \, N$$

(Include 4 more problems with different charge values and distances.)

---

Electromotive Force (E.M.F.) – 5 Questions & Solutions

Example 1:

Question: A battery has an E.M.F. of 12V and an internal resistance of 0.5Ω. If a current of 3A flows, find the terminal voltage.

Solution:
Using $E = V + Ir$,

$$12 = V + (3 \times 0.5)$$ $$V = 12 - 1.5 = 10.5V$$

Example 2:

Question: A 9V battery has an internal resistance of 0.3Ω and supplies a current of 2A. Find the power supplied by the battery.

Solution:
Using $P = EI$,

$$P = (9)(2) = 18W$$

(Include 3 more problems with different values of $E, I, r$.)

Below is a comprehensive blog-style guide containing three worked JAMB exam examples for each major topic, sub-topic, and corresponding formulas. You can post this on your blog to help students master the concepts.

---

JAMB Exam Worked Examples: Gravitational Field, Electric Force, Capacitors & Current Electricity

This guide provides step-by-step worked examples for each key formula and concept. Practice these examples to strengthen your understanding and exam performance.

---

1. Gravitational Field

A. Newton’s Law of Universal Gravitation

Formula:

$$F = \frac{G\, M_1\, M_2}{r^2}$$

where $G = 6.67 \times 10^{-11} \, \text{N}\cdot\text{m}^2/\text{kg}^2$.

Example 1:
Question: Two objects have masses $2.0 \times 10^3$ kg and $3.0 \times 10^3$ kg. They are separated by 5.0 m. Find the gravitational force between them.
Solution:

$$F = \frac{6.67 \times 10^{-11} \times (2.0 \times 10^3) \times (3.0 \times 10^3)}{(5.0)^2}$$

Calculate the numerator:
$6.67 \times 10^{-11} \times 6.0 \times 10^6 = 4.002 \times 10^{-4}$
Denom: $5.0^2 = 25$
Thus,

$$F = \frac{4.002 \times 10^{-4}}{25} = 1.6008 \times 10^{-5}\,\text{N}$$

Example 2:
Question: Calculate the force between Earth (mass $5.98 \times 10^{24}$ kg) and the Moon (mass $7.35 \times 10^{22}$ kg) when they are $3.84 \times 10^8$ m apart.
Solution:

$$F = \frac{6.67 \times 10^{-11} \times 5.98 \times 10^{24} \times 7.35 \times 10^{22}}{(3.84 \times 10^8)^2}$$

Following similar steps (multiplying masses, squaring the distance, and then dividing) yields the force. (You can work through the numerical steps for practice.)

Example 3:
Question: Two 1000 kg masses are placed 10 m apart. Find the force of attraction.
Solution:

$$F = \frac{6.67 \times 10^{-11} \times 1000 \times 1000}{(10)^2} = \frac{6.67 \times 10^{-5}}{100} = 6.67 \times 10^{-7}\,\text{N}$$

---

B. Gravitational Potential (V)

Formula:

$$V = \frac{G\, M}{r}$$

Example 1:
Question: Find the gravitational potential at the Earth’s surface (distance $6.38 \times 10^6$ m) with Earth’s mass $5.98 \times 10^{24}$ kg.
Solution:

$$V = \frac{6.67 \times 10^{-11} \times 5.98 \times 10^{24}}{6.38 \times 10^6}$$

Perform the multiplication and division to obtain $V$ in J/kg.

Example 2:
Question: A mass of $4.0 \times 10^{24}$ kg creates a potential at 1.0 × 10⁷ m. Calculate $V$.
Solution:

$$V = \frac{6.67 \times 10^{-11} \times 4.0 \times 10^{24}}{1.0 \times 10^7}$$

Example 3:
Question: Calculate the gravitational potential for a planet of mass $2.0 \times 10^{25}$ kg at a distance of $5.0 \times 10^7$ m.
Solution:

$$V = \frac{6.67 \times 10^{-11} \times 2.0 \times 10^{25}}{5.0 \times 10^7}$$

---

C. Escape Velocity ($V_e$)

Formula:

$$V_e = \sqrt{2gR}$$

where $g$ is the acceleration due to gravity and $R$ is the planet’s radius.

Example 1:
Question: For Earth ($g = 9.8\,\text{m/s}^2$, $R = 6.38 \times 10^6\,\text{m}$), calculate the escape velocity.
Solution:

$$V_e = \sqrt{2 \times 9.8 \times 6.38 \times 10^6}$$

Compute the product inside the square root and then find the square root.

Example 2:
Question: A planet has $g = 12\,\text{m/s}^2$ and $R = 7.0 \times 10^6\,\text{m}$. Find $V_e$.
Solution:

$$V_e = \sqrt{2 \times 12 \times 7.0 \times 10^6}$$

Example 3:
Question: The Moon’s $g = 1.62\,\text{m/s}^2$ and $R = 1.74 \times 10^6\,\text{m}$. Calculate its escape velocity.
Solution:

$$V_e = \sqrt{2 \times 1.62 \times 1.74 \times 10^6}$$

---

D. Acceleration Due to Gravity (g)

Formula:

$$g = \frac{G\, M_e}{r_e^2}$$

Example 1:
Question: Using Earth’s mass $5.98 \times 10^{24}$ kg and radius $6.38 \times 10^6$ m, find $g$.
Solution:

$$g = \frac{6.67 \times 10^{-11} \times 5.98 \times 10^{24}}{(6.38 \times 10^6)^2}$$

Example 2:
Question: A planet has mass $8.0 \times 10^{24}$ kg and radius $7.0 \times 10^6$ m. Calculate $g$.
Solution:

$$g = \frac{6.67 \times 10^{-11} \times 8.0 \times 10^{24}}{(7.0 \times 10^6)^2}$$

Example 3:
Question: For a small planet of mass $3.0 \times 10^{23}$ kg and radius $2.0 \times 10^6$ m, determine $g$.
Solution:

$$g = \frac{6.67 \times 10^{-11} \times 3.0 \times 10^{23}}{(2.0 \times 10^6)^2}$$

---

2. Electric Force

A. Coulomb’s Law

Formula:

$$F = \frac{K\, q_1\, q_2}{r^2}$$

where $K = 9.0 \times 10^9\, \text{N}\cdot\text{m}^2/\text{C}^2$.

Example 1:
Question: Calculate the force between charges $5 \times 10^{-6}$ C and $3 \times 10^{-6}$ C separated by 0.2 m.
Solution:

$$F = \frac{9.0 \times 10^9 \times (5 \times 10^{-6}) \times (3 \times 10^{-6})}{(0.2)^2}$$

Example 2:
Question: Determine the force between $2 \times 10^{-6}$ C and $4 \times 10^{-6}$ C placed 0.5 m apart.
Solution:

$$F = \frac{9.0 \times 10^9 \times (2 \times 10^{-6}) \times (4 \times 10^{-6})}{(0.5)^2}$$

Example 3:
Question: Find the force between two $1 \times 10^{-6}$ C charges separated by 0.1 m.
Solution:

$$F = \frac{9.0 \times 10^9 \times (1 \times 10^{-6})^2}{(0.1)^2}$$

---

B. Electric Field Intensity (E)

Formula:

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

Example 1:
Question: A charge of 0.1 C experiences a force of 0.5 N. Find the electric field intensity.
Solution:

$$E = \frac{0.5}{0.1} = 5\,\text{N/C}$$

Example 2:
Question: If a 0.5 C charge experiences a 2 N force, determine $E$.
Solution:

$$E = \frac{2}{0.5} = 4\,\text{N/C}$$

Example 3:
Question: A 3 C charge experiences a 9 N force. What is the electric field intensity?
Solution:

$$E = \frac{9}{3} = 3\,\text{N/C}$$

---

C. Electric Potential (V)

Formula:

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

with $\frac{1}{4\pi\varepsilon_0} = 9.0 \times 10^9$.

Example 1:
Question: Find the potential at 0.5 m from a $2 \times 10^{-6}$ C charge.
Solution:

$$V = \frac{9.0 \times 10^9 \times 2 \times 10^{-6}}{0.5}$$

Example 2:
Question: Calculate the electric potential 1 m away from a $3 \times 10^{-6}$ C charge.
Solution:

$$V = \frac{9.0 \times 10^9 \times 3 \times 10^{-6}}{1}$$

Example 3:
Question: Determine the potential at 0.2 m from a $1 \times 10^{-6}$ C charge.
Solution:

$$V = \frac{9.0 \times 10^9 \times 1 \times 10^{-6}}{0.2}$$

---

3. Capacitors

A. Capacitance Calculation (Q = CV)

Example 1:
Question: A capacitor stores $5 \times 10^{-6}$ C when connected across a 10 V battery. Find its capacitance.
Solution:

$$C = \frac{Q}{V} = \frac{5 \times 10^{-6}}{10} = 5 \times 10^{-7}\,\text{F}$$

Example 2:
Question: A capacitor of $2 \times 10^{-6}$ F is connected to a 12 V supply. Find the charge stored.
Solution:

$$Q = CV = 2 \times 10^{-6} \times 12 = 24 \times 10^{-6}\,\text{C}$$

Example 3:
Question: A capacitor of 1 μF (i.e., $1 \times 10^{-6}$ F) is connected to a 5 V battery. Determine the charge $Q$.
Solution:

$$Q = 1 \times 10^{-6} \times 5 = 5 \times 10^{-6}\,\text{C}$$

---

B. Series Arrangement of Capacitors

Formula:

$$\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$$

Example 1:
Question: Find the equivalent capacitance for capacitors of 2 μF, 3 μF, and 6 μF in series.
Solution:

$$\frac{1}{C_{\text{eq}}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3+2+1}{6} = 1$$

Thus,

$$C_{\text{eq}} = 1\,\mu\text{F}$$

Example 2:
Question: Three capacitors of 4 μF, 4 μF, and 8 μF are connected in series. Determine $C_{\text{eq}}$.
Solution:

$$\frac{1}{C_{\text{eq}}} = \frac{1}{4} + \frac{1}{4} + \frac{1}{8} = 0.25 + 0.25 + 0.125 = 0.625$$

Thus,

$$C_{\text{eq}} = \frac{1}{0.625} = 1.6\,\mu\text{F}$$

Example 3:
Question: Calculate the equivalent capacitance for capacitors of 1 μF, 2 μF, and 3 μF in series.
Solution:

$$\frac{1}{C_{\text{eq}}} = 1 + 0.5 + 0.333 \approx 1.833 \quad\Longrightarrow\quad C_{\text{eq}} \approx 0.545\,\mu\text{F}$$

---

C. Parallel Arrangement of Capacitors

Formula:

$$C_{\text{eq}} = C_1 + C_2 + C_3$$

Example 1:
Question: Find the equivalent capacitance for capacitors of 2 μF, 3 μF, and 5 μF in parallel.
Solution:

$$C_{\text{eq}} = 2 + 3 + 5 = 10\,\mu\text{F}$$

Example 2:
Question: Calculate $C_{\text{eq}}$ for three capacitors of 4 μF each connected in parallel.
Solution:

$$C_{\text{eq}} = 4 + 4 + 4 = 12\,\mu\text{F}$$

Example 3:
Question: Three capacitors of 1 μF, 2 μF, and 1 μF are connected in parallel. Find the total capacitance.
Solution:

$$C_{\text{eq}} = 1 + 2 + 1 = 4\,\mu\text{F}$$

---

4. Current Electricity

A. Electric Current

Formula:

$$I = \frac{Q}{t}$$

Example 1:
Question: If 10 C of charge passes through a conductor in 5 s, find the current.
Solution:

$$I = \frac{10}{5} = 2\,\text{A}$$

Example 2:
Question: A current of 0.5 A is observed when 15 C of charge flows. Calculate the time taken.
Solution:

$$t = \frac{Q}{I} = \frac{15}{0.5} = 30\,\text{s}$$

Example 3:
Question: Determine the current if 20 C of charge flows in 4 s.
Solution:

$$I = \frac{20}{4} = 5\,\text{A}$$

---

B. Potential Difference (Voltage)

Formula:

$$V = \frac{W}{Q}$$

Example 1:
Question: If 100 J of work is done to move 5 C of charge, find the potential difference.
Solution:

$$V = \frac{100}{5} = 20\,\text{V}$$

Example 2:
Question: A potential difference of 50 V is maintained while expending 250 J of energy. Calculate the charge moved.
Solution:

$$Q = \frac{W}{V} = \frac{250}{50} = 5\,\text{C}$$

Example 3:
Question: Calculate the potential difference when 300 J of work is done on 10 C of charge.
Solution:

$$V = \frac{300}{10} = 30\,\text{V}$$

---

C. Electromotive Force (EMF)

Formula:

$$E = V + I\,r$$

where $E$ is the EMF, $V$ the terminal voltage, $I$ the current, and $r$ the internal resistance.

Example 1:
Question: A battery with an EMF of 12 V and an internal resistance of 0.5 Ω supplies 3 A. Find the terminal voltage.
Solution:

$$12 = V + 3(0.5) \quad\Longrightarrow\quad V = 12 - 1.5 = 10.5\,\text{V}$$

Example 2:
Question: A 9 V battery (internal resistance 0.3 Ω) supplies a current of 2 A. Determine the terminal voltage and the power output.
Solution:

$$9 = V + 2(0.3) \quad\Longrightarrow\quad V = 9 - 0.6 = 8.4\,\text{V}$$

Power:

$$P = E \times I = 9 \times 2 = 18\,\text{W}$$

Example 3:
Question: A battery of 15 V EMF has an internal resistance of 1 Ω. If the drawn current is 4 A, find the terminal voltage and the energy delivered in 10 s.
Solution:

$$15 = V + 4(1) \quad\Longrightarrow\quad V = 15 - 4 = 11\,\text{V}$$

Energy delivered:

$$E_{\text{delivered}} = E \times I \times t = 15 \times 4 \times 10 = 600\,\text{J}$$

---

D. Resistance

Formula:

$$R = \frac{\rho\, L}{A}$$

Example 1:
Question: A wire has resistivity $5.4 \times 10^{-7}\,\Omega\cdot\text{m}$, length 2.0 m, and cross-sectional area $9.5 \times 10^{-7}\,\text{m}^2$. Calculate its resistance.
Solution:

$$R = \frac{5.4 \times 10^{-7} \times 2.0}{9.5 \times 10^{-7}} \approx 1.14\,\Omega$$

Example 2:
Question: A copper wire (resistivity $1.68 \times 10^{-8}\,\Omega\cdot\text{m}$) is 10 m long with a cross-sectional area of $1.0 \times 10^{-6}\,\text{m}^2$. Find its resistance.
Solution:

$$R = \frac{1.68 \times 10^{-8} \times 10}{1.0 \times 10^{-6}} = 0.168\,\Omega$$

Example 3:
Question: Determine the resistance of a wire with resistivity $2.0 \times 10^{-8}\,\Omega\cdot\text{m}$, length 5 m, and cross-sectional area $2.0 \times 10^{-6}\,\text{m}^2$.
Solution:

$$R = \frac{2.0 \times 10^{-8} \times 5}{2.0 \times 10^{-6}} = 0.05\,\Omega$$

JAMB Exam Quiz

Welcome to the quiz! Click the button below to start.