Exams › JEE Main › Physics

Two capacitors, each having capacitance 40 μF are connected in series. The space between one of the capacitors is filled with dielectric material of dielectric constant K such that the equivalence capacitance of the system became 24 μF. The value of K will be:

1. 1.5
2. 2.5
3. 1.2
4. 3

Correct answer: 1.5

Solution

When capacitors are connected in series, the total capacitance can be calculated using the formula for series capacitance. By substituting the values of the individual capacitors and the effect of the dielectric material, we find that the dielectric constant K must be 1.5 to achieve the equivalent capacitance of 24 μF.

Related JEE Main Physics questions

• If an electric field vector E exists in a material whose dielectric constant is K, and ε0 denotes the permittivity of free space, which expression gives the electric displacement vector?
• If a unit positive charge is taken from point A to point B along the same equipotential surface, then what can be said about the potential difference V_A - V_B?
• Two thin metallic spherical shells are placed concentrically with radii R1 and R2, where R1 > R2. If the shells carry charges Q1 and Q2 respectively, what is the electric potential at a point located at distance r such that R2 < r < R1? (Take k = 1/(4πϵ0))
• An alpha particle is made to move through a potential difference of 10⁶ V. What kinetic energy does it gain?
• A parallel-plate capacitor has its entire gap filled with a dielectric whose dielectric constant changes with position x according to K(x) = K0 + λx, where λ is a constant. How is its capacitance C related to the vacuum capacitance C0?
• Two thin circular rings of radius R each are kept on the same axis with their centres separated by a distance d. One ring carries charge +q and the other carries charge −q. What is the potential difference between the centres of the two rings?

⚔️ Practice JEE Main Physics free + battle 1v1 →