   Chapter 16, Problem 35P

Chapter
Section
Textbook Problem

a parallel-plate capacitor with area 0.200 m2 and plate separation of 3.00 mm is connected to a 6.00-V battery. (a) What is the capacitance? (b) How much charge is stored on the plates? (c) What is the electric field between the plates? (d) Find the magnitude of the charge density on each plate. (e) Without disconnecting the battery, the plates are moved farther apart. Qualitatively, what happens to each of the previous answers?

(a)

To determine

The capacitance.

Explanation

Given info: The distance between the plates (d) is 3.00 mm. The area of the plate is 0.200m2. The applied potential difference is 6.00 V.

Formula to calculate the capacitance is,

• ε0 is the permittivity of free space.
• A is the area of plate.

Substitute 0.200m2 for A, 3.00 mm for d and 8.85×1012 C2/Nm2 for ε0.

C=(8.85×1012 C2/Nm2)(0

(b)

To determine

The charge on each plate.

(c)

To determine

The electric field between the plates.

(d)

To determine

The charge density on each plate.

(e)

To determine

The effect of increasing the distance of separation between the plates.

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