A parallel plate capacitor has charge 5×10⁻⁶ C. A dielectric slab is inserted between the plates and almost fills the space between the plates. If the induced charge on one face of the slab is 4×10⁻⁶ C, then the dielectric constant of the slab is .

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5. Inside a dielectric the field is reduced to (E₀)/K, and the bound (induced) charge that produces this reduction is q'=q(1-1/K). Substituting the data: 4×10⁻⁶=5×10⁻⁶(1-1/K). 1-1/K=4/5=0.8, so 1/K=0.2. K=5.

A parallel plate capacitor has charge 5×10⁻⁶ C. A dielectric slab is inserted between the plates and almost fills the space between the plates. If the induced charge on one face of the slab is 4×10⁻⁶ C, then the dielectric constant of the slab is ______.

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