A parallel plate capacitor has charge 5 10^-6C. 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^-…

Question

A parallel plate capacitor has charge 5 10^-6C. 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^-6C then the dielectric constant of the slab is .

Accepted Answer

Correct answer: 5. (5) To find the dielectric constant K of the slab, we use the formula for the induced charge Q_ in in a parallel plate capacitor: Q_ in=Q(1- 1K) Given: Total charge Q=5 10^-6C Induced charge Q_ in=4 10^-6C We substitute the values into the formula: 4 10^-6=5 10^-6(1- 1K) To solve for the dielectric constant K, simplify the equation: Divide both sides by 5 10^-6 : 45=1- 1K Rearrange to solve for 1K : 1- 45= 1K 15= 1K Invert both sides to find K : K=5 Therefore, the dielectric constant of the slab is K=5

$A parallel plate capacitor has charge 5 \times 10^{- 6} C . A dielectric slab is inserted between the plates and almost fills the space between the plates. If the induced charge on one face of the slab is 4 \times 10^{- 6} C then the dielectric constant of the slab is .$

Solution

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