Question

I don't know how to approach this question...

Any dielectric material other than vacuum has a maximum electric field that can be produced in the dielectric material before it physically or chemically breaks down and begins to conduct. This maximum electric field is called dielectric strength. The dielectric strength for a particular material is reached at a value of 3.2 x 107 V/m. Calculate the maximum charge that can be placed on the capacitor of plate separation 0.6 cm and of C = 37 pF at this dielectric strength.

Answer 1

Qmax = C * Vmax = C*d*(V/d)max
= C*d*(dielectric strength)
= 37*10^-12 F *0.006 m * 3.2*10^7 V/m
= 7.1 *10^-6 Coulombs
since 1 Farad*Volt = 1 Coulomb

Answer 2

How did you go from
C * Vmax
to
C*d*(V/d)max
to
C*d*(dielectric strength)

Answer 3

drwls

in equation
Q max = c *Vmax = C*D* (V/d)max

did you just divide and multiply C*V with d so that you end up with dielectric strength (V/d)

Answer 4

In going from
C * Vmax
to
C*d*(V/d)max
I just multiplied and divided by d at the same time (no change) and put one of the d's under the Vmax

Vmax/d is the electric strength, in volts per meter.

Perhaps you were confused by my writing it as (V/d)max . It means the same thing. For a given d, there is a maximum V.

Answer 5

Oh okay I understand now, you just made it equivalent to get E = V/d

thanks