Here the capacitance, C=dAkϵ0 CK1,CK2,CK3 are in parallel combination and they are connected in series with CK4
Here, CK1= d/2(A/3)K1ϵ0=32 K1C, Ck2= d/2(A/3)k2ϵ0=32k2C CK3= d/2(A/3)k3ϵ032k3CC CK4= d/2(A)k4ϵ0=2k4C Now the equivalent capacitance for the combination of four capacitors is Ccq1=(Ck1+Ck2+Ck3)1+Ck41Ceq=kC kC1=2C3[k1+k2+k31]+2k4C1 or k1=2(k1+k2+k3)3+2(k4)1 or k=32(k1+k2+k3)+2(k4)
A parallel-plate capacitor of area A, plate separation d and capacitance C is filled with four dielectric materials having dielectric constantsk_1,…
NEET UG 2016 — Physics Electromagnetism
A parallel-plate capacitor of area A, plate separation d and capacitance C is filled with four dielectric materials having dielectric constantsk1, k2, k3 and k4 as shown in the figure below. If a single dielectric material is to be used to have the same capacitance C in this capacitor, then its dielectric constant k is given by

Held on 30 Apr 2016 · Verified 9 Jul 2026.
Options
- A
k=k1+k2+k3+3k4
- B
k=32(k1+k2+k3)+2k4
- C
k2=k1+k2+k33+k41
- D
k=32k4(k1+k4k1+k2+k4k2+k3+k4k3)
Did you get this right?
Sign in to track your attempts and accuracy.
Your note
Sign in to keep a private note on this question. Nothing you write is ever public.
NEET UG Electromagnetism in other years
More NEET UG Electromagnetism questions
Two point charges +2μC and -2μC are placed 10 cm apart. The electric field at the midpoint is:
A model for quantized motion of an electron in a uniform magnetic field $B$ states that the flux passing through the orbit of the electron is $n(h / e)$ where $n$ is an integer, $h$ is Planck's constant and $e$ is the magnitude of electron's charge. According to the model, the magnetic moment of an electron in its lowest energy state will be ( $m$ is the mass of the electron)
A 2 amp current is flowing through two different small circular copper coils having radii ratio $1: 2$. The ratio of their respective magnetic moments will be
Two identical charged conducting spheres $A$ and $B$ have their centres separated by a certain distance. Charge on each sphere is q and the force of repulsion between them is $F$. A third identical uncharged conducting sphere is brought in contact with sphere A first and then with B and finally removed from both. New force of repulsion between spheres $A$ and $B$ (Radii of $A$ and $B$ are negligible compared to the distance of separation so that for calculating force between them they can be considered as point charges) is best given as :
An electron (mass $9 \times 10^{-31} \mathrm{~kg}$ and charge $1.6 \times 10^{-19} \mathrm{C}$ ) moving with speed $\mathrm{c} / 100(\mathrm{c}=$ speed of light) is injected into a magnetic field $\vec{B}$ of magnitude $9 \times 10^{-4} \mathrm{~T}$ perpendicular to its direction of motion. We wish to apply an uniform electric field $\vec{E}$ together with the magnetic field so that the electron does not deflect from its path. Then (speed of light $\mathrm{c}=3 \times 10^8 \mathrm{~ms}^{-1}$ )
Other Physics topics for NEET UG
Keep practicing Electromagnetism
Work through every NEET UG Electromagnetism PYQ, year by year.