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V vs D ... potential vs distance??
seems linear ... that's quite unusual ... since electrostatic forces are inverse square force.
let's try using ratio's
V is constant from 2 to 4 means E=0 at that region?
V6/V5 = r5/r6 ???
yes ... E is derived form V by differentiating with respect to r
So what is E at d=5?
are you sure it's a staight line??
it shouldn't be zero at 6
thats quite unusual ... we usually have hyperbolic or ... what do we call this |dw:1333575004239:dw| type of graph.
Yeah but here this is the potential caused by many charges perhaps
let's see what it could be ... if it's linear that the answer is average between 4 and 6, but still it's worth it to see what causes such potential
huh how is it the average?
And yeah perhaps there is an electric field by a certain group of charges that produce this V
since the graph is linear ..
We need E
V/d = E
d = 1 ... mostlikely we have object in the costant V
That was my first step.But that gives the answer as E=1
Answer is 2.5 V/m
V = 2.5 d=1 ... so my hunch was right
both from graph
Oh no i misread it
V = average on 4 and 6, which will give 5/2
Here he gave another similar one.The electric potential decreases uniformly from 120 V to 80 V as one moves on the x axis from x=-1 cm to x=+1 c.Find E at origin.btw how much would you rate the difficulty of the questions I asked out of 5
That is V
was it rated V out of V??
I didnt get you
just forget it .. it was nothing
What is the answe rto this one
Need to find E not V
wait ... what is the answer??
E = dV/dr
E = -dV/dr