Battery Charger!

Electricity and Magnetism Level 2

Calculate the voltage of a battery connected to a parallel plate capacitor with a plate area of 2 cm 2 2\text{ cm}^2 and the plate separation of 2 mm 2\text{ mm} if the charge stored on the plates is 4.0 pC 4.0\text{ pC} .

3 0 10 5 4.5

Anish Harsha by Anish Harsha , 20, India

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1 solution

Assuming sufficient time has passed so that both the battery and the parallel plate capacitor are at the same potential (and additionally assuming that the battery has no internal resistance) ,

We have,

V b a t t e r y = V c a p a c i t o r = C h a r g e C a p a c i t a n c e \displaystyle V_{battery} = V_{capacitor} = \frac{Charge}{Capacitance}

Now,

C a p a c i t a n c e , C = A r e a ∗ ε 0 d p l a t e s = ( 2 ∗ 1 0 − 4 ) ∗ ( 8.85 ∗ 1 0 − 12 ) ( 2 ∗ 1 0 − 3 ) = 0.885 p F \displaystyle Capacitance, C = \frac{Area * ε_0}{d_{plates}} = \frac{(2*10^{-4})*(8.85*10^{-12})}{(2*10^{-3})} = 0.885 pF

Thus, V b a t t e r y = 4 ( p C ) 0.885 ( p F ) = 4.5 V \displaystyle V_{battery} = \frac{4 (pC)}{0.885 (pF)} = \boxed{4.5 V}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

You could use short names, like A for Area. Whatever, thumbs up for your precise solution!

Muhammad Arifur Rahman - 5 years, 3 months ago

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Thanks! I will keep that in mind! Cheers :)

B.S.Bharath Sai Guhan - 5 years, 3 months ago

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Most of people answer level 2 problems briefly. But you provided the best detailed solution, that's appreciable.

Muhammad Arifur Rahman - 5 years, 3 months ago

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