1.

Two parallel plates (each of area 200cm2)of a capacitor are given equal and opposite charges of magnitude 1.78×10^-6C.The space between the plates is filled with a dielectric.if dielectric field in the dielectric is 1.5×10^6V/m.What will the magnitude of charge induced on each dielectric plate ?

Answer»

We know formula,
\bf{\sigma_{ind}=\sigma\left(<klux>1</klux>-\frac{1}{K}\right)}

where \sigma_{ind} is induced surface CHARGE density on each dielectric.
\sigma is initial surface charge on each PLATE.
K is dielectric CONSTANT.

we ALSO know , Q = \sigma A

so, \frac{Q_{ind}}{A}=\frac{Q}{A}\left(1-\frac{1}{K}\right)

= Q_{ind}=Q\left(1-\frac{1}{K}\right)

now, put Q = 1.78 × 10^-6 C , K = 1.5 × 10^6

so, Qind = 1.78 × 10^-6(1 - 1/1.5 × 10^6)
= 1.7799988 × 10^-6 C


Discussion

No Comment Found

Related InterviewSolutions