Question

A parallel plate capacitor of plate area 600 and plate separation 2 mm is connected to a dc source of 200v calculate the magnitude of uniform electric feild e between the plates and the charge density on any plate

Answer 1

C= 8.85 × 10^-12 × 600 / 2 × 10-³
C= 8.85 × 10^-9 × 300
C= 26.55 F

E = 200 / 2 × 10^-2
E = 10⁴

Charge density = E × €
= 8.85 × 10^-8

Answer 2

Answer:

Given:

Area a = 600 cm2

Plate separation d = 2.0 mm

D.C. Source = 200 v

To Find:

The magnitude of the uniform electric field e between the plates.

Explanation:

$C =$$8.85\times10x^{-12} \times\frac{600}{2}\times10x^{-3}$

$C =$ $8.85\times10^{9} \times300$

C =$C = 26.55F$

$E = \frac{200}{2} \times10x^{-2}$

E = 10⁴

Charge density = E × €

=  $8.85 \times 10x^{-8}$

Hence, The magnitude of the uniform electric field e between the plates and the charge density on any plate are

E = 10⁴

Charge Density  =   $8.85 \times10x^{-8}$

Additional Information:

A uniform electric field is a field in which the value of the field strength remains the same at all points. In a uniform electric field, as the field strength does not change and the field lines tend to be parallel and equidistant to each other. They are equally spaced.

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