Question

Two 1.5 cm -diameter disks face each other, 2.4 mm apart. They are charged to ±18nC. What is the electric field strength between the disks?

Answer 1

`E = 1.15 * 10^7 N/C`

Step-by-step explanation:

As we know that disc can be treated as sheet of uniform charge density if the electric field is to be calculated at a point near its surface

so we can say that here electric field between two opposite charged disc is given as

`E = (\sigma)/(2\epsilon_0) + (\sigma)/(2\epsilon_0)`

so we have

`E = (\sigma)/(\epsilon_0)`

here we know that

`\sigma = (Q)/(\pi R^2)`

`\sigma = (18* 10^(-9))/(\pi((0.015)/(2))^2)`

`\sigma = 1.02* 10^(-4) C/m^2`

now we have

`E = (1.02* 10^(-4))/(8.85 * 10^(-12))`

`E = 1.15 * 10^7 N/C`