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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.

Respuesta :

Answer:

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 = [tex]\dfrac{K A \epsilon_0}{d}[/tex]

             [tex]\C =\dfrac{3.4\times 87.4 \times 10^{-4} \times 8.8542 \times 10^{-12}}{0.2\times 10^{-3}}[/tex]

                  C = 1.32 × 10⁻⁹ F

Q = C V                              

Q = 1.32 × 10⁻⁹ × 30                  

Q = 3.96 × 10⁻⁸ C