With the battery connected, fill the gap by a slab with the dielectric constant 3.4. if the potential is 30 v , the plate separation is 0.2 mm , and the plate area is 87.4 cm2 , find the electric charge on the plate. the value of the permittivity of free space is 8.8542 × 10−12 c 2 /n · m2 . answer in units of c.

Q = 3.96 × 10⁻⁸ C

Explanation:

given,

dielectric constant (K) = 3.4        

potential = 30 V        

plate separation = 0.2 mm      

Plate area = 87.4 cm²          

Capacitance =

                  C = 1.32 × 10⁻⁹ F

Q = C V                              

Q = 1.32 × 10⁻⁹ × 30                  

Q = 3.96 × 10⁻⁸ C

the components of a vector are the parts of the vectors that are to each other.

b perpendicular

about 45° and at the middel