A parallel-plate capacitor, having a capacitance
of 1.3 × 10^11 F and a plate spacing of
1.22 cm, is charged to a potential of 522V.
What is the electric field between the plates
if a slab of dielectric constant 1.41 is inserted
into the capacitor?
Answer in units of V/m
of 1.3 × 10^11 F and a plate spacing of
1.22 cm, is charged to a potential of 522V.
What is the electric field between the plates
if a slab of dielectric constant 1.41 is inserted
into the capacitor?
Answer in units of V/m
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Electric field between the plates is Charge density/permitivity of free space if there is no dielectric in between. If you insert a dielectric, the field will be partially canceled out by a factor of the dielectric constant. So divide the original field by the dielectric constant
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the electric field between the plates should be zero. because E fields are vectors which cancel out.