Force between two charges in medium vs vacuum — compare

Question

Two point charges $q_1 = 4 \times 10^{-6}$ C and $q_2 = -2 \times 10^{-6}$ C are placed 0.3 m apart.

(a) Calculate the force between them in vacuum.
(b) Calculate the force if they are placed in a medium with dielectric constant (relative permittivity) $K = 5$ .
(c) Explain why the force is different and define dielectric constant physically.

Solution — Step by Step

Coulomb’s Law gives the force between two point charges in vacuum (free space):

F_{vacuum} = \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 (Coulomb’s constant, $k_e$ )
$\varepsilon_0 = 8.85 \times 10^{-12}$ F/m (permittivity of free space)
$r$ = separation distance

F_{vacuum} = 9 \times 10^9 \times \frac{|4 \times 10^{-6} \times (-2 \times 10^{-6})|}{(0.3)^2}

= 9 \times 10^9 \times \frac{8 \times 10^{-12}}{0.09}

= 9 \times 10^9 \times \frac{8 \times 10^{-12}}{9 \times 10^{-2}}

= 9 \times 10^9 \times \frac{8}{9} \times 10^{-10}

= 8 \times 10^{-1} = 0.8 \text{ N}

Force in vacuum = 0.8 N (attractive, since charges are of opposite sign)

In a medium with dielectric constant $K$ (relative permittivity $\varepsilon_r$ ), the permittivity increases to $\varepsilon = K\varepsilon_0$ :

F_{medium} = \frac{1}{4\pi\varepsilon} \cdot \frac{|q_1 q_2|}{r^2} = \frac{1}{4\pi K\varepsilon_0} \cdot \frac{|q_1 q_2|}{r^2} = \frac{F_{vacuum}}{K}

The force in a medium is reduced by the factor $K$ .

F_{medium} = \frac{F_{vacuum}}{K} = \frac{0.8}{5} = 0.16 \text{ N}

Force in medium = 0.16 N (attractive)

The force is 5 times weaker in the medium than in vacuum.

The dielectric constant $K$ (or relative permittivity $\varepsilon_r$ ) is a dimensionless number that tells us how much a medium reduces the electrostatic force compared to vacuum.

Physically: When an electric field is applied to a dielectric medium, the polar molecules align partially with the field, and non-polar molecules develop induced dipoles. These aligned dipoles create an internal electric field that opposes the applied field. The net field inside the medium is weaker → force between charges is reduced.

$K = 1$ : vacuum (no reduction)
$K = 2$ : force halved
$K = 80$ : water (force reduced to 1/80 of vacuum value — this is why ions dissolve easily in water)

Definition: $K = \varepsilon_r = \frac{\varepsilon}{\varepsilon_0} = \frac{F_{vacuum}}{F_{medium}}$ (ratio of force in vacuum to force in medium at same charge-separation)

Why This Works

The dielectric medium “shields” the charges from each other. The medium’s molecules, when polarised by the charges’ electric fields, partially neutralise those fields — reducing the effective force. This is the same reason capacitors filled with dielectric materials store more charge: the dielectric reduces the effective electric field between the plates, allowing more charge to accumulate.

In JEE Main, a common question type: “Force between charges is $F$ in vacuum. If they are placed in a medium of dielectric constant $K$ , what is the new force?” Answer: $F/K$ . No calculation needed — direct formula. Similarly: “What separation in medium gives the same force as separation $r$ in vacuum?” Answer: $r/\sqrt{K}$ (since force $\propto 1/r^2$ and in medium $F_{medium} = F_{vacuum}/K$ ; setting equal gives $r_{medium} = r/\sqrt{K}$ ).

Common Mistake

Students often multiply $F_{vacuum}$ by $K$ instead of dividing. Adding a medium (which reduces the effective permittivity interaction) REDUCES the force — $F_{medium} = F_{vacuum}/K$ . Introducing a medium makes the charges “feel” each other less, not more. If you write $F_{medium} = K \times F_{vacuum} = 5 \times 0.8 = 4$ N, you’ve made this error — the answer is 5 times too large. Remember: $K > 1$ always means force decreases in the medium.

Question

Solution — Step by Step

Why This Works

Common Mistake

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