Question

If the electric field intensity between two plates is initially E₁. What would the finalelectric field intensity between parallel plates be if we double the distancebetween the plates, and half charges on the plates, and triple the area of theplatesO a. 3E₁O b. 12 E₁OC. E₁/3O d. E₁/6

Answer 1

Given:

The electric field intensity between the parallel plates is,

`E_1`

The distance between the plates is doubled and the charges are halved.

The area of the plates is tripled.

To find:

the new electric field

Step-by-step explanation:

The electric field intensity between the parallel plates is,

`E_1=(Q)/(A\epsilon)`

Q is the charge on each plate and A is the area of each plate.

Now the field intensity is,

`\begin{gathered} E_2=((Q)/(2))/(3A*\epsilon) \\ =(Q)/(6A\epsilon) \\ =(E_1)/(6) \end{gathered}`

Hence, the new electric field intensity is d.