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Electric Charges and Fields - Coulomb’s Law

Grade 12CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Coulomb’s Law states that the electrostatic force F⃗\vec{F} between two point charges q1q_1 and q2q_2 at rest is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance rr between them.

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The force acts along the line joining the two charges. It is attractive for unlike charges and repulsive for like charges.

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The constant of proportionality kk in SI units is given by k=14πϵ0k = \frac{1}{4\pi\epsilon_0}, where ϵ0\epsilon_0 is the permittivity of free space.

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The value of ϵ0\epsilon_0 is approximately 8.854×10−12 C2 N−1 m−28.854 \times 10^{-12} \text{ C}^2 \text{ N}^{-1} \text{ m}^{-2}, and k≈9×109 N m2 C−2k \approx 9 \times 10^9 \text{ N m}^2 \text{ C}^{-2}.

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In a medium with dielectric constant KK (or relative permittivity ϵr\epsilon_r), the electrostatic force decreases and is given by Fm=FvacKF_m = \frac{F_{vac}}{K}.

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Principle of Superposition: The total force on any charge due to a number of other charges is the vector sum of all the forces on that charge due to the other charges, taken one at a time.

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Coulomb's Law is valid only for point charges and for distances greater than nuclear dimensions (>10−15 m> 10^{-15} \text{ m}).

📐Formulae

F=k∣q1q2∣r2F = k \frac{|q_1 q_2|}{r^2}

k=14πϵ0=9×109 N m2 C−2k = \frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \text{ N m}^2 \text{ C}^{-2}

F⃗12=14πϵ0q1q2r122r^12\vec{F}_{12} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r_{12}^2} \hat{r}_{12}

Fmedium=14πϵ0Kq1q2r2F_{medium} = \frac{1}{4\pi\epsilon_0 K} \frac{q_1 q_2}{r^2}

ϵr=K=ϵϵ0\epsilon_r = K = \frac{\epsilon}{\epsilon_0}

F⃗net=F⃗1+F⃗2+F⃗3+...+F⃗n=∑i=1nF⃗i\vec{F}_{net} = \vec{F}_1 + \vec{F}_2 + \vec{F}_3 + ... + \vec{F}_n = \sum_{i=1}^{n} \vec{F}_i

💡Examples

Problem 1:

Two point charges q1=+2μCq_1 = +2 \mu\text{C} and q2=+6μCq_2 = +6 \mu\text{C} repel each other with a force of 12 N12 \text{ N}. If a charge of −4μC-4 \mu\text{C} is added to each of them, what will be the new force between them at the same distance?

Solution:

Initial charges: q1=2μCq_1 = 2 \mu\text{C}, q2=6μCq_2 = 6 \mu\text{C}. Force F1=12 NF_1 = 12 \text{ N}. After adding −4μC-4 \mu\text{C}, new charges are: q1′=2−4=−2μCq'_1 = 2 - 4 = -2 \mu\text{C} and q2′=6−4=+2μCq'_2 = 6 - 4 = +2 \mu\text{C}. Since F∝q1q2F \propto q_1 q_2, we have F2F1=∣q1′q2′∣∣q1q2∣\frac{F_2}{F_1} = \frac{|q'_1 q'_2|}{|q_1 q_2|}. Substituting values: F212=∣−2×2∣∣2×6∣=412\frac{F_2}{12} = \frac{|-2 \times 2|}{|2 \times 6|} = \frac{4}{12}. Thus, F2=4 NF_2 = 4 \text{ N}.

Explanation:

The magnitude of the force is proportional to the product of the charges. When the charges change from (2,6)(2, 6) to (−2,2)(-2, 2), the product of magnitudes changes from 1212 to 44. Since the product is reduced to one-third, the force is also reduced to one-third. The negative sign on one charge indicates the new force is attractive.

Problem 2:

Calculate the distance between two protons if the electrical force of repulsion between them is equal to the weight of a proton. (Mass of proton mp=1.67×10−27 kgm_p = 1.67 \times 10^{-27} \text{ kg}, Charge e=1.6×10−19 Ce = 1.6 \times 10^{-19} \text{ C})

Solution:

Force of repulsion Fe=ke2r2F_e = \frac{k e^2}{r^2}. Weight of proton W=mpgW = m_p g. Given Fe=WF_e = W, so ke2r2=mpg\frac{k e^2}{r^2} = m_p g. Solving for rr: r=ke2mpgr = \sqrt{\frac{k e^2}{m_p g}}. Substituting values: r=9×109×(1.6×10−19)21.67×10−27×9.8≈0.118 mr = \sqrt{\frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{1.67 \times 10^{-27} \times 9.8}} \approx 0.118 \text{ m}.

Explanation:

By equating the electrostatic force formula to the gravitational weight (mgmg), we can solve for the unknown distance rr. This shows the relative strength of electrostatic forces compared to gravity at the subatomic scale.

Coulomb’s Law - Revision Notes & Key Formulas | CBSE Class 12 Physics