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the potential as a function of position x is shown in the graph...

Physics
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|dw:1332350001701:dw|
i tried to draw a smooth curve
is the electric field zero at x=0? is the max magnitude at 15 or 5cm?

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Other answers:

Electric Field is defined as \[E=-\frac{dv}{dx}\] v= potential so E is the negative of the slope of the potential curve. Do note that E found from this relation is the E in x - direction. At x=0 the curve is parallel to x -axis, that is it's constant so E=0 At x=5 the slope is negative and magnitude wise the highest. So E will be max at x=5 cm
how do you determine that the curve is parallel to x axis @ x=0? it doesn't look like it on my graph but in the actual graph, it looks parallel at x=10 too
i assumed E was zero at x=0 since d(v=0) should be zero so E=(d0)/dx=0
No if v=0 doesn't mean that E=0 but the negative of slope of V is E, Yeah at x=10 also the slope is zero. So there also E is zero
hmm still confused. i need the negative of the slope of V to find where E is zero? or to find the max magnitude? and which direction is the E at the max point?
|dw:1332352015313:dw| is this what the graph of E look like?
Ok Elica, Yeah Electric field is the negative of the voltage's slope. When we have Voltage constant or parallel to axis. Then it's slope is zero and consequently E is zero
ok i understand. so is the graph i did for E right?
If V is cos then E will be sine
x=0, v=constant so E will be zero
ok so it's not right. how do i determine the direction of E at x=0 or x=10?
|dw:1332352349621:dw|
E is in the x direction always:)
so right either way
Yeah, when E is negative that means it's in -x direction
thank you so much
Welcome, Did you understand?
yes, that was great help

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