⚡ Chapter 15: Electrostatics – Numerical Problems
Prepared by Muhammad Tayyab, Subject Specialist Physics, Govt Christian High School Daska. Based on PECTAA 2026 syllabus (National Curriculum 2023).
📖 What's Inside: This section covers numerical problems from Chapter 15 Electrostatics including Coulomb's law, electric field intensity, force between charges, and charge calculations. Each problem is presented with Given Data, To Find, and step-by-step Solution as per the official PECTAA 2026 Physics curriculum. Perfect for Punjab Boards (Lahore, Gujranwala, Multan, etc.) and all BISE boards across Pakistan.
📚 Related Resources – Chapter 15: Electrostatics
Electrostatics covers Coulomb's law, electric field, potential, capacitance, charging by induction, electroscope, lightning, corona discharge, and applications. Includes solved examples and numerical problems.
📐 Numerical Problems & Solutions (PECTAA 2026)
15.1 Electric Field Intensity from Force and Charge
A positive test charge of 25 μC is placed in an electric field. The force on it is 0.500 N. What is the magnitude of the electric field at the location of the test charge?
Given Data
\(Test\ charge\)\(= q_0 = 25\ \mu C = 25 \times 10^{-6}\ C\)
\(Force\)\(= F = 0.500\ N\)
To Find
\(Electric\ field\ intensity = E = ?\)
Solution
By using formula of electric field intensity:
\[\begin{aligned}
E &= \frac{F}{q_0} \\
E &= \frac{0.500}{25 \times 10^{-6}} \\
E &= 20000\ \text{NC}^{-1} \\
\boldsymbol{E} &= \boldsymbol{2.0 \times 10^4\ \text{NC}^{-1}}
\end{aligned}\]
✅ \(E = 2.0 \times 10^4\ \text{NC}^{-1}\)
15.2 Coulomb's Force Between Two Charges
Two point charges, \(q_1 = 8\ \mu C\) and \(q_2 = 4\ \mu C\), are placed at a distance of 120 cm. What will be the Coulomb's force between them? Also, find the nature of the force.
Given Data
\(Charge\ 1\)\(= q_1 = 8\ \mu C = 8 \times 10^{-6}\ C\)
\(Charge\ 2\)\(= q_2 = 4\ \mu C = 4 \times 10^{-6}\ C\)
\(Distance\)\(= r = 120\ cm = 1.2\ m\)
To Find
\(Coulomb's\ force = F = ?\)
\(Nature\ of\ force = ?\)
Solution
By using formula of Coulomb's law:
\[\begin{aligned}
F &= k\frac{q_1 q_2}{r^2} \\
F &= \frac{(9 \times 10^9)(8 \times 10^{-6})(4 \times 10^{-6})}{(1.2)^2} \\
F &= \frac{0.288}{1.44} \\
\boldsymbol{F} &= \boldsymbol{0.2\ N}
\end{aligned}\]
Since both charges are positive, the force is repulsive.
✅ \(F = 0.2\ N\) (Repulsive)
15.3 Force Between Identical Charges at Different Distances
Two identical charges repel each other with a force of 0.2 N when they are 9 cm apart. Find the force between them when they are 3 cm apart.
Given Data
\(Initial\ force\)\(= F_1 = 0.2\ N\)
\(Initial\ distance\)\(= r_1 = 9\ cm = 0.09\ m\)
\(Final\ distance\)\(= r_2 = 3\ cm = 0.03\ m\)
To Find
\(Final\ force = F_2 = ?\)
Solution
By using formula of Coulomb's law:
\[\begin{aligned}
F_1 &= \frac{kq^2}{r_1^2} \quad \text{and} \quad F_2 = \frac{kq^2}{r_2^2} \\
\frac{F_2}{F_1} &= \frac{r_1^2}{r_2^2} \\
F_2 &= F_1 \frac{r_1^2}{r_2^2} \\
F_2 &= \frac{(0.2)(0.09)^2}{(0.03)^2} \\
F_2 &= \frac{1.62 \times 10^{-3}}{9 \times 10^{-4}} \\
\boldsymbol{F_2} &= \boldsymbol{1.8\ N}
\end{aligned}\]
✅ \(F_2 = 1.8\ N\)
15.4 Electric Field Strength from Point Charge
A point charge \(q = 10\ \mu C\) is placed in air. Calculate the electric field strength at a distance of 0.5 metres from the charge.
Given Data
\(Charge\)\(= q = 10\ \mu C = 10 \times 10^{-6}\ C\)
\(Distance\)\(= r = 0.5\ m\)
To Find
\(Electric\ field\ strength = E = ?\)
Solution
By using formula of electric field:
\[\begin{aligned}
E &= k\frac{q}{r^2} \\
E &= \frac{(9 \times 10^9)(10 \times 10^{-6})}{(0.5)^2} \\
E &= \frac{90000}{0.25} \\
E &= 360000\ \text{NC}^{-1} \\
\boldsymbol{E} &= \boldsymbol{3.6 \times 10^5\ \text{NC}^{-1}}
\end{aligned}\]
✅ \(E = 3.6 \times 10^5\ \text{NC}^{-1}\)
15.5 Repulsive Force Between Two Equally Charged Spheres
Two small equally charged metal spheres having charge \(3\ \mu C\) are brought close together such that the distance between them is 2.0 cm. Calculate the magnitude of repulsive force that each sphere exerts on the other.
Given Data
\(Charge\ on\ sphere\ 1\)\(= q_1 = 3\ \mu C = 3 \times 10^{-6}\ C\)
\(Charge\ on\ sphere\ 2\)\(= q_2 = 3\ \mu C = 3 \times 10^{-6}\ C\)
\(Distance\)\(= r = 2\ cm = 0.02\ m\)
To Find
\(Repulsive\ force = F = ?\)
Solution
By using formula of Coulomb's law:
\[\begin{aligned}
F &= k\frac{q_1 q_2}{r^2} \\
F &= \frac{(9 \times 10^9)(3 \times 10^{-6})(3 \times 10^{-6})}{(0.02)^2} \\
F &= \frac{0.081}{4 \times 10^{-4}} \\
\boldsymbol{F} &= \boldsymbol{202.5\ N}
\end{aligned}\]
✅ \(F = 202.5\ N\)
15.6 Charge on Each Sphere from Force and Distance
Two small equally charged metal spheres repel each other with a force of 0.5 N when placed 3 cm apart in the air. Calculate the charge on each sphere.
Given Data
\(Force\)\(= F = 0.5\ N\)
\(Distance\)\(= r = 3\ cm = 0.03\ m\)
To Find
\(Charge\ on\ each\ sphere = q = ?\)
Solution
By using formula of Coulomb's law:
\[\begin{aligned}
F &= k\frac{q^2}{r^2} \\
q^2 &= \frac{F r^2}{k} \\
q^2 &= \frac{(0.5)(0.03)^2}{9 \times 10^9} \\
q^2 &= \frac{4.5 \times 10^{-4}}{9 \times 10^9} \\
q^2 &= 5 \times 10^{-14} \\
q &= 2.24 \times 10^{-7}\ C \\
\boldsymbol{q} &= \boldsymbol{0.224\ \mu C}
\end{aligned}\]
Note: The textbook displays 224 μC. This appears to be a common textbook typographical error where the decimal point was omitted. The mathematically precise answer is 0.224 μC.
✅ \(q = 0.224\ \mu C\)
15.7 Finding the Other Charge from Force and Distance
A force of 85 N acts between two charges placed 3.2 cm apart. If one charge is 25 μC, determine the other charge.
Given Data
\(Force\)\(= F = 85\ N\)
\(Distance\)\(= r = 3.2\ cm = 0.032\ m\)
\(Charge\ 1\)\(= q_1 = 25\ \mu C = 25 \times 10^{-6}\ C\)
To Find
\(Charge\ 2 = q_2 = ?\)
Solution
By using formula of Coulomb's law:
\[\begin{aligned}
F &= k\frac{q_1 q_2}{r^2} \\
q_2 &= \frac{F r^2}{k q_1} \\
q_2 &= \frac{(85)(0.032)^2}{(9 \times 10^9)(25 \times 10^{-6})} \\
q_2 &= \frac{0.08704}{225000} \\
q_2 &= 3.87 \times 10^{-7}\ C \\
\boldsymbol{q_2} &= \boldsymbol{0.387\ \mu C}
\end{aligned}\]
Note: The textbook displays 3.87 μC. This appears to be a common textbook typographical error where the decimal point was omitted. The mathematically precise answer is 0.387 μC.
✅ \(q_2 = 0.387\ \mu C\)
📘 Examples with Solutions
Ex 15.1 Electric Field from Point Charge
A point charge \(q = 10\ \mu C\) is placed in air. Calculate the electric field at a distance of 0.5 m from the charge.
Given Data
\(Charge\)\(= q = 10\ \mu C = 10 \times 10^{-6}\ C\)
\(Distance\)\(= r = 0.5\ m\)
To Find
\(Electric\ field = E = ?\)
Solution
By using formula of electric field:
\[\begin{aligned}
E &= k\frac{q}{r^2} \\
E &= \frac{(9 \times 10^9)(10 \times 10^{-6})}{(0.5)^2} \\
E &= \frac{90000}{0.25} \\
E &= 360000\ \text{NC}^{-1} \\
\boldsymbol{E} &= \boldsymbol{3.6 \times 10^5\ \text{NC}^{-1}}
\end{aligned}\]
✅ \(E = 3.6 \times 10^5\ \text{NC}^{-1}\)
📐 Key Formulas – Electrostatics
Coulomb's Law: \( F = k \frac{q_1 q_2}{r^2} \)
Electric Field Intensity: \( E = \frac{F}{q_0} \)
Electric Field (Point Charge): \( E = k \frac{q}{r^2} \)
Electric Potential: \( V = \frac{W}{q} \)
Coulomb's Constant: \( k = 9 \times 10^9 \, \text{Nm}^2\text{C}^{-2} \)
Quantization of Charge: \( q = ne \)
Unit 10Thermal Physics
Unit 11Transfer of Thermal Energy
Unit 12Waves
Unit 13Sound
Unit 14Light
Unit 15✨ Electrostatics
Unit 16Electricity
Unit 17Electromagnetism
Unit 18Electromagnetic Induction & EM Waves
Unit 19Electronics
Unit 20Atomic and Nuclear Physics
Unit 21Space and Environment
📖 Complete syllabus coverage for Class 10 Physics (PECTAA 2026) – Units 10 to 21
💡 Exam Tip:
For numerical problems, always write Given Data, To Find, and step-by-step Solution with formulas. Pay attention to units and conversions. These problems follow the PECTAA 2026 pattern and are prepared by Subject Specialist Muhammad Tayyab.
Created by Hira Science Academy | Aligned with PECTAA 2026 Syllabus