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kolezko [41]
3 years ago
8

A parallel plate capacitor has two plates has a separation d. If I charge it with a +Q and a ????Q on the two plates, respective

ly, what is the electrostatic energy? What is it if I move to 1/2 d?
Physics
1 answer:
EleoNora [17]3 years ago
5 0

Answer:

Explanation:

Electric energy per unit volume

= (1 / 2) x ε₀ E²

Total energy

= energy per unit volume x volume

=  (1 / 2) x ε₀ E² x A d  , A is plate area

= (1 / 2A) x (ε₀ (σ/ε₀)² x A² d) [ E is electric field ]

(1 / 2A) x (ε₀ (σA/ε₀)² X d)

(1 / 2A) x (ε₀ (Q/ε₀)² X d

= (1 / 2A) x (Q²/ε₀) x d

So total energy is proportional to d , distance between plates

if  d is changed to d/2

total energy also becomes half.

(1 / 2A) x (Q²/ε₀) x d / 2

(1 / 4A) x (Q²/ε₀) x d

=

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