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Chapter 17: Problem 65

A 6.2 -cm by 2.2 -cm parallel plate capacitor has the plates separated by a distance of \(2.0 \mathrm{mm} .\) (a) When \(4.0 \times 10^{-11} \mathrm{C}\) of charge is placed on this capacitor, what is the electric field between the plates? (b) If a dielectric with dielectric constant of 5.5 is placed between the plates while the charge on the capacitor stays the same, what is the electric field in the dielectric?

Short Answer

Expert verified
Solution: (a) Without a dielectric, the electric field between the plates is \(3.334 \times 10^3 N/C\). (b) With a dielectric of dielectric constant 5.5, the electric field in the dielectric is \(606.182 N/C\).

Step by step solution

01

Part (a) - Electric field between the plates without the dielectric

First, we need to find the surface area of the parallel plate capacitor. Area (A) = length × width = 6.2 cm × 2.2 cm = 62 mm × 22 mm = \(1.364 \times 10^{-3} m^2\) (Converting cm to m) Now we will find the electric field between the plates using the formula: \(E = \frac{Q}{A \epsilon_0}\) where E is the electric field, Q is the charge on the capacitor, A is the area of the plates, and \(\epsilon_0\) is the vacuum permittivity (\(8.854 \times 10^{-12} C^2/Nm^2\)). Plugging in the values: \(E = \frac{4.0 \times 10^{-11} C}{1.364 \times 10^{-3} m^2 \times 8.854 \times 10^{-12} C^2/Nm^2} = 3.334 \times 10^3 N/C\) So, the electric field between the plates is \(3.334 \times 10^3 N/C\).
02

Part (b) - Electric field in the dielectric

Now we will find the electric field in the dielectric. When a dielectric is inserted between the plates of a capacitor, the electric field is reduced by a factor of the dielectric constant (K). So the formula to find the electric field in the dielectric is: \(E' = \frac{E}{K}\) where E' is the electric field in the dielectric, E is the electric field without the dielectric, and K is the dielectric constant. Plugging in the values: \(E' = \frac{3.334 \times 10^3 N/C}{5.5} = 606.182 N/C\) So, the electric field in the dielectric is \(606.182 N/C\).

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Most popular questions from this chapter

A spherical conductor with a radius of \(75.0 \mathrm{cm}\) has an electric field of magnitude \(8.40 \times 10^{5} \mathrm{V} / \mathrm{m}\) just outside its surface. What is the electric potential just outside the surface, assuming the potential is zero far away from the conductor?
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