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Grade: 12th pass

                        

(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.

4 months ago

Answers : (2)

Arun
24742 Points
							
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
4 months ago
Vikas TU
12285 Points
							
Dear student 
The above explanation is not properly explained 
Please ignore this 
We are sorry for inconvenience.
Good Luck 
Cheers 
3 months ago
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