Aparallel-plate capacitor is formed from two 1.0 cm times 1.0 cm electrodes spaced 2.9 mm apart. the electric field strength inside the capacitor is 1.0 times10^{6} n/c. what is the charge (in nc) on positive electrode?

0.89 nC

Explanation:

The strength of the electric field inside a parallel plate capacitor is given by

(1)

where

Q is the charge stored on one plate

A is the area of one plate

is the vacuum permittivity

For this problem, we have

is the electric field strength

the area of one plate is

Solving the formula (1) for Q, we find the charge on the positive electrode:

answer: decibel

explanation:

explanation:

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