Page 8 - PRE-U STPM PHYSICS TERM 2

Physics Term 2 STPM Chapter 13 Capacitors

7. On the other hand, if an air capacitor after being charged is disconnected from the battery, the
insertion of an insulator between the plates
1
(a) reduces the electric field to — its initial value,

ε
r 1
(b) reduces the potential difference to — its initial value,
ε
r
Q
(c) Since C = — , when V decreases for the same charge Q on either plates, the capacitance C
V
13 increases by a factor ε . 13
r
Example 2
(a) A parallel-plate capacitor consists of two metal plates each of area 2.0 m separated by a distance
2
4
of 5.0 mm in air. A potential difference of 1.0 × 10 V is applied across the capacitor. Calculate
(i) the capacitance,
(ii) the charge on each plate,
(iii) the electric field strength between the plates.

(ε for air = 1.00)
r
(b) The charged capacitor is then disconnected from the charging voltage and insulated so that
the charge in the capacitor is constant. A piece of dielectric of thickness 5.0 mm and dielectric
constant 5.0 is inserted in between the plates. Calculate

(i) the electric field strength between the plates,
(ii) the potential difference across the capacitor,
(iii) the capacitance.
Solution:

ε A
0
(a) (i) Capacitance, C = —– Exam Tips
d
–12
8.85 × 10 × 2 1. When a charged capacitor is disconnected from the
= ———————
5 × 10 –3 charging voltage, the charge Q remains constant.
= 3.54 × 10 F 2. If the capacitor remained connected to the
–9
charging voltage, the potential difference V remains
(ii) Charge on each plate unchanged.
Q = CV
= (3.54 × 10 ) × (1 × 10 )
–9
4
= 3.54 × 10 C
–5

(iii) Electric field strength
V
E = —
d
1 × 10 4
= —–——
5 × 10 –3
= 2.0 × 10 V m –1
6
(b) (i) Electric field strength

E
E’ = —
ε
r
2.0 × 10 6
= ————–
5
= 4.0 × 10 V m –1
5
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13 Physic T2.indd 48 10/18/18 3:18 PM