Potential difference between two points?

AI Thread Summary
The discussion revolves around calculating the potential difference between two points in an electric field created by parallel charged plates. Given that the electric potential difference across the plates is 160V and they are 8cm apart, the electric field intensity is determined to be 2000N/C. The potential difference between points A and B, which are 2cm from the positive plate and 3cm from the negative plate respectively, can be calculated using the voltage gradient of 20 volts per centimeter. The distance between points A and B is 3cm, leading to a potential difference of 60 volts. The approach emphasizes that direct calculations can simplify the process without needing to reference voltages at each point independently.
bigbird24
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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=\DeltaV/d = 160/0.08 =2000N/C
\DeltaVa =E*d ---What value for d do i use?
\DeltaVb =E*d ---What value for d do i use?
\DeltaVab = \DeltaVb - \DeltaVa
 
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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
<br /> E = d \Delta V<br />
straight from a to b?
 
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.
 
thanks AJ, that makes sense to me
 
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