Question

Calculate the intensity of electric field at a point 25 cm from a charge of 4.8uC in a medium of dielectric constant 3.6.

(a) 19.2x10²NC
(b) 192X10²N/C
(c) 192X10²N/C
(d) 1.92X10²N/C​

Answer 1

The intensity of the electric field at a point 25 cm from a charge of 4.8 μC in a medium of dielectric constant 3.6, is
`1920*10^2\ N/C` (option C)

How to calculate the intensity of electric field?

The intensity of the electric field at a point 25 cm from a charge of 4.8 μC can be calculated as illustrated below:

• Distance apart (r) = 25 cm = 25 / 100 = 0.25 m
• Charge (Q) = 4 μC =
  `4*10^(-6)\ C`
• Vacuum permittivity constant (
  `\epsilon_0`) =
  `8.854*10^(-12)\ C^2/Nm^2`
• Dielectric constant (k) = 3.6
• Intensity of electric field (E) =?

`E = (Q)/(4\pi \epsilon_0kr^2) \\\\E = (4.8*10^(-6))/(4\ *\ 3.14\ *\ 8.854*10^(-12)\ *\ 3.6\ *\ 0.25^2)\\\\E = 1920*10^2\ N/C`

From the above calculation the intensity of the electric field is calculated to be
`1920*10^2\ N/C`.

Therefore, the correct answer to the question is option C

Complete question:

Calculate the intensity of electric field at a point 25 cm from a charge of 4.8uC in a medium of dielectric constant 3.6.

(a) 19.2x10²NC

(b) 192X10²N/C

(c) 1920X10²N/C

(d) 1.92X10²N/C​