The dielectric strength of rutile is 6.0 × 106 V/m, which corresponds to the maximum electric field that the dielectric can sustain before breakdown. What is the maximum charge that a 10−10-F capacitor with a 1.08-mm thickness of rutile can hold?

6.48*10⁻⁷ C

Explanation:

By definition, the capacitance of a capacitor is expressed as follows:

       

where Q is the charge on one of the plates of the capacitor, and V the potential difference between the plates.The maximum electric field, the potential difference, and the distance between plates are related by the following expression:

       

Replacing by the givens, we can find V as follows:

       

Now, we can find the maximum charge Qmax,  as follows:

       

The maximum charge that the capacitor can hold is 6.48*10⁻⁷ C.

He can play two games. 13/4=3.253.25*2=6.5

Sorry i don’t know what to do