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Understanding Human Behaviour Without Spoken Words — Edwin Ogie Library Understanding Human Behaviour Without Spoken Words Nonverbal Communication as a core human skill — simple, practical, and classroom-friendly. Chapter Objectives Introduction Meaning & Scope Major Channels Interpreting Behaviour Culture & Ethics Practical Applications Case Illustrations Reflection & Practice Summary & Terms By Edwin Ogie Library — clear, usable lessons for students and teachers. Chapter Objectives At the end of this chapter, the reader should be able to: Clearly define nonverbal communication and explain its role in human interaction. Identify and interpret major forms of nonverbal behaviour with accuracy. Analyse behaviour using clusters of cues rather than isolated signals. Apply nonverbal awareness eff...
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Gravitational Field, Electric Force, and Current Electricit

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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=GM1M2r2F = \frac{G M_1 M_2}{r^2}

where:

  • M1,M2M_1, M_2 = Masses of the objects (kg)
  • rr = Distance between masses (m)
  • GG = Gravitational constant (6.67×10−11 N⋅m2kg−26.67 \times 10^{-11} \, \text{N} \cdot \text{m}^2 \text{kg}^{-2})

Electric Force (Coulomb’s Law)

F=Kq1q2r2F = \frac{K q_1 q_2}{r^2}

where:

  • q1,q2q_1, q_2 = Charges (C)
  • rr = Distance between charges (m)
  • KK = Coulomb’s constant (9.0×109 N⋅m2C−29.0 \times 10^9 \, \text{N} \cdot \text{m}^2 \text{C}^{-2})

Electromotive Force (E.M.F.)

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

E=V+IrE = V + Ir

where:

  • EE = Electromotive force (V)
  • VV = Terminal voltage (V)
  • II = Current (A)
  • rr = Internal resistance of the source (Ω\Omega)

Power Output of a Battery

P=EIP = EI

where:

  • PP = Power (W)
  • II = Current (A)

Worked Examples

Gravitational Field – 5 Questions & Solutions

Example 1:

Question: The mass of the Earth is 5.98×10245.98 \times 10^{24} kg, and its radius is 6.38×1066.38 \times 10^6 m. Calculate the acceleration due to gravity on the Earth's surface.

Solution:
Using g=GMere2g = \frac{G M_e}{r_e^2},

g=(6.67×10−11)(5.98×1024)(6.38×106)2g = \frac{(6.67 \times 10^{-11}) (5.98 \times 10^{24})}{(6.38 \times 10^6)^2} g=9.8 m/s2g = 9.8 \, \text{m/s}^2

(Include 4 more similar problems with varying values.)

Electric Force – 5 Questions & Solutions

Example 1:

Question: Two charges, 5×10−6C5 \times 10^{-6} C and 3×10−6C3 \times 10^{-6} C, are 0.2 m apart. Calculate the electric force between them.

Solution:
Using F=Kq1q2r2F = \frac{K q_1 q_2}{r^2},

F=(9.0×109)(5×10−6)(3×10−6)(0.2)2F = \frac{(9.0 \times 10^9) (5 \times 10^{-6}) (3 \times 10^{-6})}{(0.2)^2} F=3.375 NF = 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+IrE = V + Ir,

12=V+(3×0.5)12 = V + (3 \times 0.5) V=12−1.5=10.5VV = 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=EIP = EI,

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

(Include 3 more problems with different values of E,I,rE, 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=G M1 M2r2F = \frac{G\, M_1\, M_2}{r^2}

where G=6.67×10−11 N⋅m2/kg2G = 6.67 \times 10^{-11} \, \text{N}\cdot\text{m}^2/\text{kg}^2.

Example 1:
Question: Two objects have masses 2.0×1032.0 \times 10^3 kg and 3.0×1033.0 \times 10^3 kg. They are separated by 5.0 m. Find the gravitational force between them.
Solution:

F=6.67×10−11×(2.0×103)×(3.0×103)(5.0)2F = \frac{6.67 \times 10^{-11} \times (2.0 \times 10^3) \times (3.0 \times 10^3)}{(5.0)^2}

Calculate the numerator:
6.67×10−11×6.0×106=4.002×10−46.67 \times 10^{-11} \times 6.0 \times 10^6 = 4.002 \times 10^{-4}
Denom: 5.02=255.0^2 = 25
Thus,

F=4.002×10−425=1.6008×10−5 NF = \frac{4.002 \times 10^{-4}}{25} = 1.6008 \times 10^{-5}\,\text{N}

Example 2:
Question: Calculate the force between Earth (mass 5.98×10245.98 \times 10^{24} kg) and the Moon (mass 7.35×10227.35 \times 10^{22} kg) when they are 3.84×1083.84 \times 10^8 m apart.
Solution:

F=6.67×10−11×5.98×1024×7.35×1022(3.84×108)2F = \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=6.67×10−11×1000×1000(10)2=6.67×10−5100=6.67×10−7 NF = \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=G MrV = \frac{G\, M}{r}

Example 1:
Question: Find the gravitational potential at the Earth’s surface (distance 6.38×1066.38 \times 10^6 m) with Earth’s mass 5.98×10245.98 \times 10^{24} kg.
Solution:

V=6.67×10−11×5.98×10246.38×106V = \frac{6.67 \times 10^{-11} \times 5.98 \times 10^{24}}{6.38 \times 10^6}

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

Example 2:
Question: A mass of 4.0×10244.0 \times 10^{24} kg creates a potential at 1.0 × 10⁷ m. Calculate VV.
Solution:

V=6.67×10−11×4.0×10241.0×107V = \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 2.0×10252.0 \times 10^{25} kg at a distance of 5.0×1075.0 \times 10^7 m.
Solution:

V=6.67×10−11×2.0×10255.0×107V = \frac{6.67 \times 10^{-11} \times 2.0 \times 10^{25}}{5.0 \times 10^7}

C. Escape Velocity (VeV_e)

Formula:

Ve=2gRV_e = \sqrt{2gR}

where gg is the acceleration due to gravity and RR is the planet’s radius.

Example 1:
Question: For Earth (g=9.8 m/s2g = 9.8\,\text{m/s}^2, R=6.38×106 mR = 6.38 \times 10^6\,\text{m}), calculate the escape velocity.
Solution:

Ve=2×9.8×6.38×106V_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 m/s2g = 12\,\text{m/s}^2 and R=7.0×106 mR = 7.0 \times 10^6\,\text{m}. Find VeV_e.
Solution:

Ve=2×12×7.0×106V_e = \sqrt{2 \times 12 \times 7.0 \times 10^6}

Example 3:
Question: The Moon’s g=1.62 m/s2g = 1.62\,\text{m/s}^2 and R=1.74×106 mR = 1.74 \times 10^6\,\text{m}. Calculate its escape velocity.
Solution:

Ve=2×1.62×1.74×106V_e = \sqrt{2 \times 1.62 \times 1.74 \times 10^6}

D. Acceleration Due to Gravity (g)

Formula:

g=G Mere2g = \frac{G\, M_e}{r_e^2}

Example 1:
Question: Using Earth’s mass 5.98×10245.98 \times 10^{24} kg and radius 6.38×1066.38 \times 10^6 m, find gg.
Solution:

g=6.67×10−11×5.98×1024(6.38×106)2g = \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 8.0×10248.0 \times 10^{24} kg and radius 7.0×1067.0 \times 10^6 m. Calculate gg.
Solution:

g=6.67×10−11×8.0×1024(7.0×106)2g = \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 3.0×10233.0 \times 10^{23} kg and radius 2.0×1062.0 \times 10^6 m, determine gg.
Solution:

g=6.67×10−11×3.0×1023(2.0×106)2g = \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=K q1 q2r2F = \frac{K\, q_1\, q_2}{r^2}

where K=9.0×109 N⋅m2/C2K = 9.0 \times 10^9\, \text{N}\cdot\text{m}^2/\text{C}^2.

Example 1:
Question: Calculate the force between charges 5×10−65 \times 10^{-6} C and 3×10−63 \times 10^{-6} C separated by 0.2 m.
Solution:

F=9.0×109×(5×10−6)×(3×10−6)(0.2)2F = \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 2×10−62 \times 10^{-6} C and 4×10−64 \times 10^{-6} C placed 0.5 m apart.
Solution:

F=9.0×109×(2×10−6)×(4×10−6)(0.5)2F = \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 1×10−61 \times 10^{-6} C charges separated by 0.1 m.
Solution:

F=9.0×109×(1×10−6)2(0.1)2F = \frac{9.0 \times 10^9 \times (1 \times 10^{-6})^2}{(0.1)^2}

B. Electric Field Intensity (E)

Formula:

E=FqE = \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=0.50.1=5 N/CE = \frac{0.5}{0.1} = 5\,\text{N/C}

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

E=20.5=4 N/CE = \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=93=3 N/CE = \frac{9}{3} = 3\,\text{N/C}

C. Electric Potential (V)

Formula:

V=14πε0QrV = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r}

with 14πε0=9.0×109\frac{1}{4\pi\varepsilon_0} = 9.0 \times 10^9.

Example 1:
Question: Find the potential at 0.5 m from a 2×10−62 \times 10^{-6} C charge.
Solution:

V=9.0×109×2×10−60.5V = \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 3×10−63 \times 10^{-6} C charge.
Solution:

V=9.0×109×3×10−61V = \frac{9.0 \times 10^9 \times 3 \times 10^{-6}}{1}

Example 3:
Question: Determine the potential at 0.2 m from a 1×10−61 \times 10^{-6} C charge.
Solution:

V=9.0×109×1×10−60.2V = \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 5×10−65 \times 10^{-6} C when connected across a 10 V battery. Find its capacitance.
Solution:

C=QV=5×10−610=5×10−7 FC = \frac{Q}{V} = \frac{5 \times 10^{-6}}{10} = 5 \times 10^{-7}\,\text{F}

Example 2:
Question: A capacitor of 2×10−62 \times 10^{-6} F is connected to a 12 V supply. Find the charge stored.
Solution:

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

Example 3:
Question: A capacitor of 1 μF (i.e., 1×10−61 \times 10^{-6} F) is connected to a 5 V battery. Determine the charge QQ.
Solution:

Q=1×10−6×5=5×10−6 CQ = 1 \times 10^{-6} \times 5 = 5 \times 10^{-6}\,\text{C}

B. Series Arrangement of Capacitors

Formula:

1Ceq=1C1+1C2+1C3\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:

1Ceq=12+13+16=3+2+16=1\frac{1}{C_{\text{eq}}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3+2+1}{6} = 1

Thus,

Ceq=1 μFC_{\text{eq}} = 1\,\mu\text{F}

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

1Ceq=14+14+18=0.25+0.25+0.125=0.625\frac{1}{C_{\text{eq}}} = \frac{1}{4} + \frac{1}{4} + \frac{1}{8} = 0.25 + 0.25 + 0.125 = 0.625

Thus,

Ceq=10.625=1.6 μFC_{\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:

1Ceq=1+0.5+0.333≈1.833⟹Ceq≈0.545 μF\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:

Ceq=C1+C2+C3C_{\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:

Ceq=2+3+5=10 μFC_{\text{eq}} = 2 + 3 + 5 = 10\,\mu\text{F}

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

Ceq=4+4+4=12 μFC_{\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:

Ceq=1+2+1=4 μFC_{\text{eq}} = 1 + 2 + 1 = 4\,\mu\text{F}

4. Current Electricity

A. Electric Current

Formula:

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

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

I=105=2 AI = \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=QI=150.5=30 st = \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=204=5 AI = \frac{20}{4} = 5\,\text{A}

B. Potential Difference (Voltage)

Formula:

V=WQV = \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=1005=20 VV = \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=WV=25050=5 CQ = \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=30010=30 VV = \frac{300}{10} = 30\,\text{V}

C. Electromotive Force (EMF)

Formula:

E=V+I rE = V + I\,r

where EE is the EMF, VV the terminal voltage, II the current, and rr 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)⟹V=12−1.5=10.5 V12 = 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)⟹V=9−0.6=8.4 V9 = V + 2(0.3) \quad\Longrightarrow\quad V = 9 - 0.6 = 8.4\,\text{V}

Power:

P=E×I=9×2=18 WP = 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)⟹V=15−4=11 V15 = V + 4(1) \quad\Longrightarrow\quad V = 15 - 4 = 11\,\text{V}

Energy delivered:

Edelivered=E×I×t=15×4×10=600 JE_{\text{delivered}} = E \times I \times t = 15 \times 4 \times 10 = 600\,\text{J}

D. Resistance

Formula:

R=ρ LAR = \frac{\rho\, L}{A}

Example 1:
Question: A wire has resistivity 5.4×10−7 Ω⋅m5.4 \times 10^{-7}\,\Omega\cdot\text{m}, length 2.0 m, and cross-sectional area 9.5×10−7 m29.5 \times 10^{-7}\,\text{m}^2. Calculate its resistance.
Solution:

R=5.4×10−7×2.09.5×10−7≈1.14 ΩR = \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 1.68×10−8 Ω⋅m1.68 \times 10^{-8}\,\Omega\cdot\text{m}) is 10 m long with a cross-sectional area of 1.0×10−6 m21.0 \times 10^{-6}\,\text{m}^2. Find its resistance.
Solution:

R=1.68×10−8×101.0×10−6=0.168 ΩR = \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 2.0×10−8 Ω⋅m2.0 \times 10^{-8}\,\Omega\cdot\text{m}, length 5 m, and cross-sectional area 2.0×10−6 m22.0 \times 10^{-6}\,\text{m}^2.
Solution:

R=2.0×10−8×52.0×10−6=0.05 ΩR = \frac{2.0 \times 10^{-8} \times 5}{2.0 \times 10^{-6}} = 0.05\,\Omega



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