please leave explanations. this is just practice. A parallel-plate, air-filled capacitor has a charge of 20.0 µC and a gap width of 0.200 mm. The potential differencebetween the plates is 400 V.1) What is the electric field in the region between the plates? (Express your answer to three significant figures.)MV/m Submit2) What is the surface charge density on the positive plate? (Express your answer to three significant figures.)µC/m² Submit

Answered: A parallel-plate, air-filled capacitor…

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ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

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please leave explanations. this is just practice.

A parallel-plate, air-filled capacitor has a charge of 20.0 µC and a gap width of 0.200 mm. The potential difference
between the plates is 400 V.
1) What is the electric field in the region between the plates? (Express your answer to three significant figures.)
MV/m Submit
2) What is the surface charge density on the positive plate? (Express your answer to three significant figures.)
µC/m² Submit

Expert Solution

Step 1 Electric field between plates

Formula of an electric field between plates of capacitor is as follows:

$E = \frac{V}{d} 1) w h e r e, V = p o t e n t i a l d i f f e r e n c e b e t w e e n p l a t e s o f c a p a c i t o r d = s e p e r a t i o n o f p a r a l l e l p l a t e s$

Values given in the question are as follow:

$q = 20 μ C = 20 \times 10^{- 6} C d = 0.200 m m = 0.2 \times 10^{- 3} m V = 400 v o l t$

Substitute known values in equation-1):

$E = \frac{400 V}{0.2 \times 10^{- 3} m} E = 2 \times 10^{6} \frac{V}{m} E = 2 \frac{M V}{m}$

Electric field between plates of capacitor is $2 \frac{M V}{m}$

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