Physics

posted by .

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 -

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 -

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

• Physics -

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)

• Physics -

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.

• Physics -

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

thanks

Similar Questions

1. Physics

How does the capacitance of a parallel plate capacitor change when its plates are moved to twice their initial distance and a slab of material with dielectric constant К = 2 is placed between the plates to replace air?
2. Physics

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 …
3. physics

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 …
4. physics

Two capacitors, identical except for the dielectric material between their plates, are connected in parallel. One has a material with a material with a dielectric constant of 2, while the other has a material with a dielectric constant …
5. physics

he question is: A spherical capacitor contains a solid spherical conductor of radius 0.5mm with a charge of 7.4 micro coulombs, surrounded by a dielectric material with er = 1.8 out to a radius of 1.2mm, then an outer spherical non-conducting …
6. Capacitance (tricky)

Two parallel plates of area 41 cm.^2 have equal but opposite charges of 2.60E-07 C. Within the dielectric material filling the space between the plates, the electric field is 5.80E+05 V./m. Find the dielectric constant of the material. …
7. Physics

A certain substance has a dielectric constant of 2.8 and a dielectric strength of 18.0 MV/m. If it is used as the dielectric material in a parallel-plate capacitor, what minimum area should the plates of the capacitor have to obtain …
8. Physics

If the electric field inside a capacitor exceeds the dielectric strength of the dielectric between its plates, the dielectric will break down, discharging and ruining the capacitor. Thus, the dielectric strength is the maximum magnitude …
9. Physics

We can physically model the axonal membrane (axolemma) as a parallel plate capacitor with dielectric in it. If the capacitance per unit area in F/m^2 is known, what single additional parameter needs to be measured in order to determine …
10. Physics

Given Two parallel plate capacitors with a seperation d(metres) in free space connected to a dc supply of volts,v. A dielectric material of thickness b(metres) is inserted between the plates so that a distance (d-b)m remains without …

More Similar Questions