Q.40 The electric field and capacitance of a capacitor in the absence of dielectric material are E and C, respectively. When the capacitor is filled with a dielectric material, the electric field and capacitance of the capacitor becomes E′ and C′, respectively. Which of the following is (are) correct?
(A) E′ > E and C′ = C
(B) E′ < E and C′ > C
(C) E′ = E and C′ > C
(D) E′ > E and C′ < C
Option (B) is correct: E′ < E and C′ > C.
When a dielectric fills a capacitor connected to a constant voltage source (typical assumption unless specified otherwise), the material polarizes and creates an opposing field that reduces the net electric field between plates, while capacitance increases due to higher effective permittivity.
Core Concepts
Capacitance C=ϵ0A/d without dielectric becomes C′=kC (where k>1 is the dielectric constant) when filled completely, increasing charge storage for the same voltage. The electric field drops from E=V/d to E′=E/k because polarization induces bound charges opposing the original field.
Option Analysis
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(A) E′ > E and C′ = C: Incorrect. Dielectric reduces field strength and always boosts capacitance.
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(B) E′ < E and C′ > C: Correct for constant voltage (battery connected). Voltage V stays fixed, so reduced E′ pairs with higher C′.
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(C) E′ = E and C′ > C: Incorrect. Field magnitude decreases; it cannot remain unchanged with dielectric polarization.
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(D) E′ > E and C′ < C: Incorrect. Neither change occurs—field drops, capacitance rises.
The effect of dielectric on capacitor electric field and capacitance transforms basic capacitor performance. In the absence of dielectric, electric field E and capacitance C define operation; with dielectric, they become E′ and C′. This guide analyzes changes, focusing on why E′ < E and C′ > C holds true.
Dielectric Polarization Mechanism
Dielectric molecules polarize in the applied field, shifting bound charges to create an internal field opposing E. Net field reduces to E′=E/k, where k (dielectric constant >1) quantifies this effect. Capacitance rises as C′=kϵ0A/d, enabling more charge storage.
Constant Voltage vs. Constant Charge Cases
| Scenario | Voltage | Charge | Electric Field | Capacitance |
|---|---|---|---|---|
| Battery Connected | Constant V | Increases (Q′=kQ) | E′<E (by 1/k) | C′>C (by k) |
| Isolated (Charged) | Decreases (V′=V/k) | Constant Q | E′<E (by 1/k) | C′>C (by k) |
In both, C′>C and E′<E, matching option (B). Questions assume constant V unless stated otherwise.
Practical Implications
Higher C′ suits energy storage devices; reduced E′ prevents breakdown. Examples: ceramic capacitors (k≈1000) vs. air (k=1).
