**Sample Problem 26-6a**
The capacitance of the plates before the dielectric slab is inserted is equal to 8.207661pF

**Sample Problem 26-6b**
Free charge on the plates is equal to701.755040pC

**Sample Problem 26-6c**
The electric field is equal to 6895.161290V/m

**Sample Problem 26-6d**
The electric field in dielectric slab is equal to 2641.824249V/m

**Sample Problem 26-6e**
The new potential difference is equal to 52.323971V

**Sample Problem 26-6f**
The new capacitance is equal to 13.411731pF