(a) Two plates (20 cm x 3.0 cm) are separated by a 1.0 mm air gap forming a capacitor. Assume dielectric constant () for air is 1.000. The capacitor is connected to a 12 V battery. i) Determine the capacitance of the parallel plate capacitor. ii) What is the charge on each plate? (b) Three particles of charge 50 C, 27 C and –70 C are placed in a straight line with the 27 C charge in the middle and the 50 C charge to its left. The charge in the middle is 0.30 m from each of the other two charges. Calculate the magnitude and direction of net force on the 27 C charge due to the other two charges.

Arun
25750 Points
3 years ago
C=
d
ϵ
o
​
A
​

According to the values given C=1.8×10
−11
F
Now Q=CV
Hence charge on the positive plate will be 1.8×10
−9
C
(a) Dielectric constant, k = 6
C=1.8×10
−11
F
new capacitance, C
=kC=108 pF
V=100volts
=V×C
=1.08×10
−8
C
V remains the same.

(b) C
=kC=6×1.8×10
−11
=108pF
q remains the same.
q=1.8×10
−9
C
=q/C
=16.67 V
Vikas TU
14149 Points
3 years ago
Dear student
The above explanation is not properly explained
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