Textbook ExpertVerified Tutor
25 Nov 2021
Given information
A capacitor of capacitance C= 250pF.
Step-by-step explanation
Step 1.
It is known that the energy of a capacitor of capacitance C is calculated as,
where V is is the potential difference.
In this problem we need to get four times the initial energy.
Therefore, final energy, 
Also according to the problem we need to get four times the energy by using an extra capacitor with C.
Thus we must add the new capacitor in parallel with C, as in parallel connection the capacitance is added.
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Physics: Principles with Applications
7th Edition, 2013
Giancoli
ISBN: 9780321625922
Solutions
Chapter
Section
- Problem 1a
- Problem 1b
- Problem 2
- Problem 3
- Problem 4
- Problem 5a
- Problem 5b
- Problem 6a
- Problem 6b
- Problem 7
- Problem 8
- Problem 10
- Problem 11
- Problem 12a
- Problem 12b
- Problem 13
- Problem 14
- Problem 15
- Problem 16a
- Problem 16b
- Problem 16c
- Problem 17
- Problem 18a
- Problem 18b
- Problem 19
- Problem 20
- Problem 21
- Problem 22a
- Problem 22b
- Problem 22c
- Problem 22d
- Problem 22e
- Problem 23
- Problem 24a
- Problem 24b
- Problem 24c
- Problem 24d
- Problem 25
- Problem 26
- Problem 27
- Problem 28a
- Problem 28b
- Problem 29a
- Problem 29b
- Problem 30
- Problem 31
- Problem 32
- Problem 33a
- Problem 33b
- Problem 34a
- Problem 34b
- Problem 35
- Problem 36
- Problem 37
- Problem 38a
- Problem 38b
- Problem 39
- Problem 40a
- Problem 40b
- Problem 41
- Problem 42
- Problem 43
- Problem 44
- Problem 45
- Problem 46a
- Problem 46b
- Problem 46c
- Problem 47
- Problem 48
- Problem 49
- Problem 50
- Problem 51a
- Problem 51b
- Problem 52
- Problem 53
- Problem 54a
- Problem 54b
- Problem 56
- Problem 57a
- Problem 57b
- Problem 57c
- Problem 58a
- Problem 58b
- Problem 58c
- Problem 58d
- Problem 59a
- Problem 59b
- Problem 60
- Problem 61
- Problem 62a
- Problem 62b
- Problem 63
- Problem 64
- Problem 65
- Problem 66
- Problem 71
- Problem 72a
- Problem 72b
- Problem 73a
- Problem 73b
- Problem 73b
- Problem 73b
- Problem 73b