Consider a large parallel plate capacitor. The gap d between the two plates is filled entirely with a dielectric slab of relative permittivity 5. The plates are initially charged to a potential difference of V volts and then disconnected from the source. If the dielectric slab is pulled out completely, then the ratio of the new electric field E2 in the gap to the original electric field E1 is ___________.
Concept:
The capacitance of a parallel plate capacitor is
\(C = \frac{{ϵ A}}{d}\)
Where ϵ = ϵoϵr is the dielectric constant.
Permittivity of free space ϵo = 8.854 × 10-14 C2/N-cm2
ϵr is the relative permittivity of the dielectric medium
ϵr = 1 for the air medium.
A is the area of cross-section of the plates
d is the distance between the plates
The electric field intensity between the plates is
E = V / d
V is the voltage across the plates.
The charge stored in the capacitor is
Q = CV = \(\frac{{ϵ_o ϵ_r AV}}{d}=ϵ_o ϵ_r AE\) ....(1)
Calculation:
Given that the plates are initially charged to a potential difference of V volts
Charge stored Q in both cases is the same.
From equation(1), the electric field intensity is inversely proportional to the relative permittivity of the medium
E ∝ 1 / ϵr
\(\frac{{E_2}}{{E_1}}=\frac{{\epsilon_{r1}}}{{\epsilon_{r2}}}\)
\(\frac{{E_2}}{{E_1}}=\frac{{5}}{{1}} = 5\)Important Questions from Electrostatics
Electric potential $V$, is a ____ field, and electric field intensity $E$, is a ______ field.
The dielectric constant of a vacuum is _____.
- What is the magnitude of the electric field at a distance $r$ from a point charge $Q$?
According to Gauss’s Law, the surface integral of the normal component of electric flux density D over a closed surface containing charge Q is:
The electric potential at the surface of an atomic nucleus (z = 50) of radius 9 × 10-15 m is ______________.
