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 3.1 x 107 V/m. Calculate the maximum charge that can be placed on the capacitor of plate separation 0.4 cm and of C = 57 pF at this dielectric strength.
i am very very confused with this question, please help.
I think this is the third time this has been posted in the past week. Have you read my previous answer? Some students keep posting under changing names.
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To calculate the maximum charge that can be placed on the capacitor, we need to use the formula:
Q = CV
Where:
Q is the charge on the capacitor
C is the capacitance of the capacitor
V is the voltage across the capacitor
In this case, the maximum electric field in the dielectric material, also known as the dielectric strength, is given as 3.1 x 10^7 V/m. The voltage across the capacitor can be calculated using the formula:
V = E × d
Where:
V is the voltage across the capacitor
E is the electric field strength
d is the plate separation distance
Given:
E = 3.1 x 10^7 V/m
d = 0.4 cm = 0.4 × 10^-2 m
Calculating V:
V = (3.1 x 10^7 V/m) × (0.4 × 10^-2 m)
V = 1.24 x 10^5 V
Now that we have the voltage, we can calculate the maximum charge on the capacitor.
Given:
C = 57 pF = 57 × 10^-12 F
Calculating Q:
Q = (1.24 x 10^5 V) × (57 × 10^-12 F)
Q = 7.068 x 10^-6 C
Therefore, the maximum charge that can be placed on the capacitor is 7.068 x 10^-6 C.