Posted by Sandra on Thursday, March 4, 2010 at 4:51am.
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.
- Physics - drwls, Thursday, March 4, 2010 at 5:50am
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
- Physics - anony mouse, Friday, March 5, 2010 at 1:10am
How did you go from
C * Vmax
- Physics - PEACE :), Saturday, March 6, 2010 at 1:09am
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)
- Physics - drwls, Sunday, March 7, 2010 at 3:49pm
In going from
C * Vmax
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.
- Physics - anony mouse, Sunday, March 7, 2010 at 5:52pm
Oh okay I understand now, you just made it equivalent to get E = V/d
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