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

Answer:

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

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

$V=2\times\text{Potential halfway between the plates}$

Put the value into the formula

$V=2\times15$

$V=30\ V$

We need to calculate the electric field between the plates

Using formula of electric field

$E=\dfrac{V}{d}$

$E=\dfrac{30}{1.6\times10^{-2}}$

$E=1875\ V/m$

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

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