ch25_p38 - × 10 − 10 C and q 2 = C 2 V 2 = 2.66 × 10...

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38. Each capacitor has 12.0 V across it, so Eq. 25-1 yields the charge values once we know C 1 and C 2 . From Eq. 25-9, C 2 = ε 0 A d = 2.21 × 10 11 F , and from Eq. 25-27, C 1 = κε 0 A d = 6.64 × 10 11 F . This leads to q 1 = C 1 V 1 = 8.00
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Unformatted text preview: × 10 − 10 C and q 2 = C 2 V 2 = 2.66 × 10 − 10 C. The addition of these gives the desired result: q tot = 1.06 × 10 − 9 C. Alternatively, the circuit could be reduced to find the q tot ....
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