Tag: electrostatic potential and capacitance

The force of attraction between two charges separated by certain distance in air is F1. If the space between the charges is completely filled with dielectric of constant 4 the force becomes F2. If half of the distance between the charges is filled with same dielectric the force between the charges is F3. Find F1:F2:F3 is

Water possesses a high dielectric constant, therefore

Two charges placed at a distance experience a force of x N.The force between the charges when each charge is doubled, distance between them is halved and charge placed in a medium of dielectric constant 2.52 is.

Two fixed charges separated by a distance $d$ experience a force $F$. A dielectric medium of thickness $\dfrac{d}{4}$ and dielectric constant $4$ is introduced in the space between them. Find the new force between the charges.

The relative permittivity of water is 81. If $\varepsilon _0$ and $\varepsilon$ are permittivities of vaccum and water respectively Then:

Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle of ${ 30 }^{ o }$ with each other. When suspended in a liquid of density $0.8g { cm }^{ -3 }$ the angle remains the same. If density of the material of the sphere is $1.6g { cm }^{ -3 }$ the dielectric constant of the liquid is:

The capacitance of an air filled parallel plate capacitor is $8 pF$. The separation between the plates is doubled and the space between the plates is then filled with wax giving the capacitance a new value of $80 \times 10^{-12}$ farads. 
The dielectric constant of wax is

The dielectric strength of air is nearly ___ V/mm

What is the value of dielectric constant for metal ?

Surface charge density of a disc is given by $\sigma = \dfrac{\sigma _o r}{R}$ where $\sigma _0$ is a constant, $r$ is distance from the center, and $R$ is the radius of the disc. If electric potential at the centre of the disc is $\dfrac{\sigma _0 R}{\alpha \epsilon _0}$ then find value of $\alpha$.