Potential difference between two points?

[bigbird24]

This is a question from my highscool gr12 physics class
Q) Two parallel plates are 8cm apart and are appositely charged. The electric potential difference across the plates is 160V. Point A is 2cm from the positive plate, point B is 3 cm from the negative plate.
-What is potential difference between points A and B

My attempt-
Electric field intensity: E=$$\Delta$$V/d = 160/0.08 =2000N/C
$$\Delta$$Va =E*d ---What value for d do i use?
$$\Delta$$Vb =E*d ---What value for d do i use?
$$\Delta$$Vab = $$\Delta$$Vb - $$\Delta$$Va

 

[zhermes]

because you're looking for potential difference you don't need to calculate the reference voltages at a and b independently. What if you just used
$$E = d \Delta V$$
straight from a to b?

 

[AJ Bentley]

You don't need formulae, you can do it in your head.

The plates are 8cm apart so at 160V there is a 20 volts per centimetre gradient between them.
The points are 3 cm apart so the answer is 60 volts.

 

[bigbird24]

thanks AJ, that makes sense to me

 

[rwisz]

= E*(d1-d2)

The potential difference between two points is a measure of the difference in electric potential energy per unit charge between those two points. It is also known as voltage and is represented by the symbol V. In this scenario, the potential difference between points A and B can be calculated by subtracting the potential at point A from the potential at point B. This can be done using the formula \DeltaV = \DeltaVb - \DeltaVa.

To calculate \DeltaVa and \DeltaVb, we can use the formula \DeltaV = E*d, where E is the electric field intensity and d is the distance between the points and the respective plates. In this case, since point A is 2cm from the positive plate and point B is 3cm from the negative plate, we can use d=0.02m for \DeltaVa and d=0.03m for \DeltaVb.

Therefore, the potential difference between points A and B, \DeltaVab, is given by:

\DeltaVab = \DeltaVb - \DeltaVa = (E*d2) - (E*d1) = E*(d2-d1)

Substituting the values, we get:

\DeltaVab = 2000N/C * (0.03m - 0.02m) = 2000N/C * 0.01m = 20V

Hence, the potential difference between points A and B is 20V. This means that for every unit of charge moving from point A to point B, it would gain 20 joules of electric potential energy. This potential difference is a result of the electric charges on the parallel plates and the distance between them.