Question

Two parallel plates, connected to a 100 V power supply, are separated by an air gap. How small can the gap be if the air is not to exceed its breakdown value of E = 3 x 10^6 V/m?

Answer 1

use the equation E=V/l where E: electric field (here breakdown field)
V: potential difference between plates
l: distance between plates (this is what u need to find)

Answer 2

DERIVATION for the above formula-----
V means potential difference i,e work done to move a unit positive charge from one point to another. E means force on a unit positive charge due to external sources . and l means distance between the 2 points ..now put them all together,,,,
work done = force X displacement
for a unit charge
V = E X l

Answer 3

Let d be the distance between the plates
Use formula : V=Ed

If you want an analysis,
find Q from Q=CV (introduce the variable A) and substitute it in -
\[E=\frac{ Q }{ Aε_{0} }\]
A will cancel out.

Answer 4

LastDayWork , your formula involves CAPACITANCE, and that is not involved in the question,,

Answer 5

Two parallel plates connected to a battery - sounds like a capacitor to me.

Answer 6

Btw, I don't think using additional theory to explain a concept is a bad approach

Answer 7

I agree that we are dealing with a capacitor,but calculation of capacitance is not possible in the given question, also AREA 'A' is not given in the above question, so we cant apply your formula.........as u mentioned , "find Q from Q=CV" ,,this cannot be done because capacitance of the capacitor isnt given,,

Answer 8

and how will A cancel out?

Answer 9

\[E=\frac{ Q }{ A \epsilon _{0} }=\frac{ CV }{ A \epsilon _{0} }=\frac{ \frac{ A \epsilon _{0} }{d } V}{ A \epsilon _{0} }\]
implies
\[E=\frac{ V }{ d }\]

Answer 10

^^ @AnthonyStark

Answer 11

oh..i see,,im really sorry,,,never thought in this way,,..,must say,,that was GENIUS.

Answer 12

Thanks :)