Coulomb's law numerical — force between two charges with dielectric medium

Question

Two point charges of $+3 \mu C$ and $-4 \mu C$ are placed 20 cm apart in vacuum. (a) Find the force between them. (b) If a dielectric medium of dielectric constant $K = 4$ is introduced between them, find the new force.

(NCERT Class 12, Chapter 1 — Electric Charges and Fields)

Solution — Step by Step

The force between two point charges in vacuum:

F = \frac{1}{4\pi\varepsilon_0} \cdot \frac{|q_1||q_2|}{r^2}

where $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9$ N m²/C².

$q_1 = 3 \mu C = 3 \times 10^{-6}$ C, $q_2 = 4 \mu C = 4 \times 10^{-6}$ C, $r = 20$ cm $= 0.2$ m

F = 9 \times 10^9 \times \frac{3 \times 10^{-6} \times 4 \times 10^{-6}}{(0.2)^2}

F = 9 \times 10^9 \times \frac{12 \times 10^{-12}}{0.04}

F = 9 \times 10^9 \times 3 \times 10^{-10}

\boxed{F = 2.7 \text{ N (attractive, since charges are opposite)}}

When a dielectric of constant $K$ fills the space between the charges, the force reduces by a factor of $K$ :

F' = \frac{F}{K} = \frac{2.7}{4}

\boxed{F' = 0.675 \text{ N}}

The dielectric weakens the electrostatic force because the polarised molecules partially cancel the field between the charges.

Why This Works

Coulomb’s law is the electrical analogue of Newton’s gravitational law — both follow the inverse square law. The key difference: gravitational force is always attractive, while electrostatic force can be attractive or repulsive depending on the signs of the charges.

A dielectric medium contains molecules that get polarised in the electric field. These induced dipoles create an opposing field inside the medium, effectively reducing the net force. The dielectric constant $K$ (also called relative permittivity $\varepsilon_r$ ) quantifies this reduction. For vacuum, $K = 1$ . For water, $K \approx 80$ — which is why ionic compounds dissolve easily in water (the inter-ionic force drops drastically).

Alternative Method — Using the formula with medium directly

F' = \frac{1}{4\pi\varepsilon_0 K} \cdot \frac{|q_1||q_2|}{r^2} = \frac{1}{4\pi\varepsilon} \cdot \frac{|q_1||q_2|}{r^2}

where $\varepsilon = K\varepsilon_0$ is the permittivity of the medium. This is equivalent to dividing the vacuum force by $K$ .

For quick NEET calculations, first compute the numerator: $q_1 \times q_2 = 12 \times 10^{-12}$ . Then compute $r^2 = 4 \times 10^{-2}$ . The ratio gives $3 \times 10^{-10}$ . Multiply by $9 \times 10^9$ to get 2.7 N. Breaking it into small steps reduces calculation errors.

Common Mistake

Two frequent errors: (1) Forgetting to convert $\mu C$ to Coulombs — $1 \mu C = 10^{-6}$ C, not $10^{-3}$ C. (2) Forgetting to convert cm to metres for the distance. NEET questions deliberately give mixed units to catch careless students. Always convert everything to SI units before substituting.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using the formula with medium directly

Common Mistake

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