Version One – Homework 1 – Juyang Huang – 24018 – Jan 16 ...

Version One – Homework 1 – Juyang Huang – 24018 – Jan 16, 2008 28
A 15 pF capacitor is connected across a 75 V
source.
What charge is stored on it
Correct answer: 1.125 × 10 −9 C.
Explanation:
Let : C = 15 pF = 1.5 × 10 −11 F and
V = 75 V .
The capacitance is
keywords:
C = q V
q = C V
= (1.5 × 10 −11 F) (75 V)
= 1.125 × 10 −9 C
Capacitance Comparison 02
26:02, trigonometry, multiple choice, > 1 min,
fixed.
056 (part 1 of 1) 10 points
A parallel plate capacitor is connected to a
battery.
5. None of these.
Explanation:
The capacitance of a parallel plate capacitor
is
C = ɛ 0
A
d .
Hence doubling d halves the capacitance,
and Q = C V is also halved
keywords:
(
C ′ = ɛ 0
A
2 d = 1 2 ɛ 0
A
d = 1 )
2 C .
Plate Separation
26:02, trigonometry, numeric, > 1 min, normal.
057 (part 1 of 1) 10 points
A parallel-plate capacitor has a plate area of
12 cm 2 and a capacitance of 7 pF.
The permittivity of a vacuum is 8.85419 ×
10 −12 C 2 /N · m 2 .
What is the plate separation
Correct answer: 0.00151786 m.
Explanation:
+Q −Q
Let : A = 12 cm 2 = 0.0012 m 2 ,
C = 7 pF = 7 × 10 −12 F , and
ɛ 0 = 8.85419 × 10 −12 C 2 /N · m 2 .
¢ £
d
¤ ¥
2 d
If we double the plate separation,
1. the capacitance is doubled.
2. the electric field is doubled.
3. the potential difference is halved.
4. the charge on each plate is halved. correct
C = ɛ 0 A
d
d = ɛ 0 A
( C
8.85419 × 10 −12 C 2 /N · m 2)
=
keywords:
7 × 10 −12 F
× ( 0.0012 m 2)
= 0.00151786 m .
AP B 1993 MC 15 16