Question

The plates of a parallel plate capacitor are separated by d = 1.6 cm. The potential of the negative plate is 0 V, and the potential halfway between the plates is +15 V (see the drawing). What is the electric field between the plates (take the upward direction as the positive direction)? Be sure to include the proper + or - sign.

Answer 1

The electric field between the plates is 1875 V/m.

Step-by-step explanation:

Given that,

Separated d=1.6 cm

Potential of negative plate = 0 V

Potential halfway between the plates = 15 V

We need to calculate the total potential

Using formula of potential

`V=2*\text{Potential halfway between the plates}`

Put the value into the formula

`V=2*15`

`V=30\ V`

We need to calculate the electric field between the plates

Using formula of electric field

`E=(V)/(d)`

`E=(30)/(1.6*10^(-2))`

`E=1875\ V/m`

Direction is negative as the field always points from positive to negative plate

Hence, The electric field between the plates is 1875 V/m.