A capacitor consists of two charged disks of radius 4.3 m separated by a distanc
ID: 1333229 • Letter: A
Question
A capacitor consists of two charged disks of radius 4.3 m separated by a distance s = 2 mm (see the figure). The magnitude of the charge on each disk is 45 mu C. Consider points A, B, C, and D inside the capacitor, as shown in the diagram. The distance si and the distance S2 = 0.6 mm. (Assume the + x axis is to the right, the + y axis is up, and the + z axis is out.) What is the electric field inside the capacitor? First, calculate the potential difference VB - VA. What is along this path? Next, calculate the potential difference Vc - Vb = What is Vc along this path? Finally, calculate the potential difference Va - Vc- What is along this path?Explanation / Answer
(a)
The electric field inside is only in the x direction and is equal to
E = q / A
= 45 x 10-6 / (8.85 x10-12 * ) (4.32 )
=87535 N/C
The electric files is,
E = (87535, 0, 0) N/C
(b)
Going from A to B,
L= ( -0.0013, 0, 0 ) m
Therefore,
VB - VA = - 87535* (-0.0013)
= 113.8 V
(c) Going from B to C,
L= (0,-0.0006, 0 ) m
Therefore,
VC - VB = - 131879 * 0
= 0 Volts
(d)
Going from C to A ,
L =( 0.0013 , 0.0006 , 0 ) m
Therefore,
VA - VC = - 131879 * 0.0013 + 0 * 0.0006
=-113.8 Volts
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