Question

Problem 2.

Compute the dielectric constants of the media in the following scenarios:

i. The electric potential due to a 1∗10−9C charge 1 meter away on a test charge is 9 volts. What might be the medium in this interaction?

ii. The electrostatic energy is 2.93∗10−20J for two electrons one angstrom apart. What might be the medium in this interaction?

Answer 1

1)

Electric Potential, V is given by

----(1)

Ο 1 V Απερε, Τ κ

Where Q is the charge, r is the distance

The value of

1 Απερ-

is 9*10⁹

^{Rearranging the equation (1) we get,}

E,=9 109 Vr

^{Substituting the above values we get,}

$\varepsilon$_r = 9*10⁹*1*10^-9/(9*1)

=> $\varepsilon$_r = 1

This is the relative permittivity of vacuum. Therefore, the medium is vacuum (also called as free space)

ii) Electrostatic energy, U(r) is given by

------(2)

Τ υr)- κ Απερε, Υ

Rearranging equation (2) we get,

E9 109 rU (r)

Substituting the value of q = Q = 1.6x10^-19 C (which is the value of charge of electron), r = 1x10^-10 m, U=2.93x10^-20 J we get,

$\varepsilon$_r = 9*10⁹*1.6*10^-19*1.6*10^-19/(1*10^-10*2.93*10^-20)

$\varepsilon$_r = 78.6

This value is the relative permittivity of pure water. Therefore the medium in this case is pure water.