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Grade 11Electric Current

a potential differance is applied btw the plates of parallel plate capacitor of area Aand seperation d,.the battery is disconnected and a dielactric of constant K is placed btw plates of capacitor of thickness t

Profile image of charitha .bvs
14 Years agoGrade 11
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1 Answer

Profile image of SAGAR SINGH - IIT DELHI
14 Years ago

Dear charitha,

C= K*Eo*A/D, where  Eo= 8.854x10-12

where:

K is the dielectric constant of the material,
A is the overlapping surface area of the plates,
d is the distance between the plates, and
C is capacitance

A (plate area) (mm2)
d (distance) (mm)
K (dielectric constant )

Capacitance (uF)
Capacitance (pF)