NEET Questions

The dimensional formula for the gravitational constant G multiplied by the permittivity of free space $\epsilon_0$ is:

The quantity $\sqrt{\frac{\epsilon_0}{\mu_0}}$ has the dimensions of:

Two charges are placed a certain distance apart in air. If a glass slab is introduced between them, the force between them will:

A spherical capacitor with inner radius $a$ and outer radius $b$ is filled with a dielectric whose permittivity varies as $\epsilon = \frac{\epsilon_0}{1 + r/a}$. What is its capacitance?

The capacitance of a parallel plate capacitor with air as medium is $6\mu F$. With the introduction of a dielectric medium, the capacitance becomes $30\mu F$. The permittivity of the medium is: ($$\epsilon_0 = 8.85 imes 10^{-12} C^2N^{-1}m^{-2}$$)

A parallel plate capacitor has a capacitance of $4 \mu F$ with air as the dielectric. A dielectric material is inserted, increasing the capacitance to $20 \mu F$. What is the relative permittivity of the dielectric?

If the capacitance of a parallel plate capacitor increases from $10 pF$ to $50 pF$ when a dielectric is introduced, what is the dielectric constant of the material?

A student constructs a parallel plate capacitor with air between the plates. Its capacitance is measured to be $2 \mu F$. Then, the student fills the space between the plates with a dielectric material, and the capacitance increases to $10 \mu F$. What is the permittivity of the dielectric material if the permittivity of free space is $\epsilon_0 = 8.85 imes 10^{-12} C^2 N^{-1} m^{-2}$?

The capacitance of a parallel plate air capacitor is $8 \mu F$. A dielectric material with a relative permittivity of 4 is inserted between the plates. What is the new capacitance?

The capacitance of a parallel plate capacitor with air as medium is $3\mu F.$ With the introduction of a dielectric medium between the plates, the capacitance becomes $15\mu F.$ The permittivity of the medium is