# Calculating electric field strength at point.Watch

#1
Two charged parallel metal plates, X and Y, are separate by a distance of 2.0m.
X is at a potential of -150V and Y is at a potential of +150V. Point P is midway between X and Y. Which of the following gives the electric field strength at point P?
A) 150 V m^-1 to the right.
B) 150 V m^-1 to the left.
C) 300 V m^-1 to the right.
D) 300 V m^-1 to the left

https://imageshack.com/i/porA5QtQj
The link has a picture of the question.
Thank you in advance! If somebody can explain why the answer is B, I would be grateful! Thanks!
0
11 months ago
#2
The electric field strength between two parallel plates is uniform - it will have the same value at every point between the plates, so the actual position of P is irrelevant (so long as it is between the plates!).

Do you know an equation for the electric field strength between two parallel plates with a potential difference of V between them, and which are separated by a perpendicular distance d? This should help you to see why the magnitude of the electric field strength is 150 V m^-1.

As for direction - electric fields are taken by convention to "point" from positive to negative, so the direction of the field is from the plate with the larger potential to that of lower potential (taking sign into account, not just magnitude). This explains why the electric field strength - which is a vector - points to the left.
1
#3
(Original post by Pangol)
The electric field strength between two parallel plates is uniform - it will have the same value at every point between the plates, so the actual position of P is irrelevant (so long as it is between the plates!).

Do you know an equation for the electric field strength between two parallel plates with a potential difference of V between them, and which are separated by a perpendicular distance d? This should help you to see why the magnitude of the electric field strength is 150 V m^-1.

As for direction - electric fields are taken by convention to "point" from positive to negative, so the direction of the field is from the plate with the larger potential to that of lower potential (taking sign into account, not just magnitude). This explains why the electric field strength - which is a vector - points to the left.
Thank you for resolving why it goes to the left and not the right. I didn't know that it went from largest potential to lowest potential.

As for the equation, the equation i think this relates to is E= V/d " The electric field strength between two oppositely charged parallel plates" https://revisionworld.com/a2-level-l...lectric-fields

When E = Electric field strength, V is potential difference between the plates and d is the separation. Is the difference between the plates 300V? and because distance is 2 metres, 300/2 = 150 V therefore the right answer? Is this how it is explained?
0
11 months ago
#4
(Original post by Kyle2114)
Thank you for resolving why it goes to the left and not the right. I didn't know that it went from largest potential to lowest potential.
From positive to negative is how it is usally expressed.

(Original post by Kyle2114)
When E = Electric field strength, V is potential difference between the plates and d is the separation. Is the difference between the plates 300V? and because distance is 2 metres, 300/2 = 150 V therefore the right answer? Is this how it is explained?
Yes, that's it. If one of the plates was at +150 V and the other at 0 V, then clearly the potential difference between them is 150 V. In this case, you have that amount of difference twice over. Or you could just look at it mathematically - what is the difference between +150 and -150?
0
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