What happens to the potential difference between the plates of a capacitor as the thickness of the d

Question

What happens to the potential difference between the plates of a capacitor as the thickness of the dielectric slab increases?

Examoogle · Accepted Answer

Decreases

Explanation: When a dielectric slab of thickness 't' is inserted into a capacitor of plate separation 'd', the net electric field inside the dielectric is reduced due to polarization. This reduction in the total potential difference across the plates leads to an increase in capacitance, keeping the charge constant.

Working Principle: The total potential difference $V$ across the plates is the sum of the potential drop across the air gap ($V_{air}$) and the dielectric ($V_{diel}$). Since $V = E_0(d-t) + \frac{E_0}{K}t$, where $K$ is the dielectric constant and $E_0$ is the field in vacuum, as $t$ increases, the term $(d-t) + \frac{t}{K}$ decreases, resulting in a decrease in $V$ for a fixed charge $Q$.

Key Formulas: $V = E_0(d - t) + \frac{E_0}{K}t$ — Potential difference with dielectric; $C = \frac{\epsilon_0 A}{d - t(1 - \frac{1}{K})}$ — Effective capacitance

Key Points: Inserting a dielectric increases the capacitance $C$.; For a constant charge $Q$, since $V = Q/C$, an increase in $C$ must result in a decrease in $V$.; The factor $(1 - \frac{1}{K})$ represents the effective reduction in distance.; If the slab fills the entire space ($t=d$), the potential becomes $V = V_0/K$.

Advantages: Increased energy storage capability; Higher dielectric strength prevents breakdown

Disadvantages: Increased complexity in manufacturing; Dielectric loss at high frequencies

Applications: Filter circuits; Energy storage devices; Coupling capacitors

Additional Info: The potential difference decreases because the polarization charges on the dielectric surface create an opposing electric field, reducing the net field between the plates.. Option A is incorrect because increasing dielectric thickness enhances capacitance, which forces the potential difference to drop for a fixed charge source.

Answer: B is correct — As the thickness of the dielectric slab increases, the effective capacitance of the capacitor increases, which, at a constant charge, leads to a reduction in the potential difference across the plates.

Exam Tip: Always remember that for capacitors, Capacitance ($C$) and Potential ($V$) are inversely proportional when charge ($Q$) is constant ($V = Q/C$).

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