More Electrostatics

[MedPR]

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A charged particle is suspended between two horizontally aligned parallel plates carrying opposite charges of equal magnitude. If the separation between the plates is doubled while the voltage is held constant, what happens to the electrostatic force on a particular drop?

A. doubled
B. halved
C. reduced to one-fourth
D. stays the same

Answer is: B


My question will give away the answer, but here it is anyway.

Why doesn't F=kqq/r^2 apply?

 

[chiddler]

Because we are not looking at two discrete charges. We are looking at the behavior of a single charge in an electric field.

So it's like asking what is F=kqq/r^2 of an electron in an electric field, with no other protons/neutrons around?

Don't know! The closest we can do is figure out the total charge of each charged plate and call them q2 and q3. But even then I don't think this is accurate because they take such a wide area of space whereas a single electron is more of a point charge, know what I mean?

 

[SaintJude]

EK stresses this! (p.119 if you got it) That formula you noted is only applicable when an electric field is created by a point charge.

If your electric field is constant (as it it in a capacitor) stay away from that formula.

 

[chiddler]

yeah st jude worded it better than me.

 

[MedPR]

Thanks again both of you.

 

[Class1P]

So what equation would you use? the one for potential energy ?

U=E*q*d

or U=F*d

so U=( k*q1*q2)/r

 

[MrNeuro]

Huh??? But the question is asking about a force using the formula will give you a j/C not a Newton.....and I looked up the EK ref u = kq/r is also included along with the other electrostatic formulas.