A capacitor consists of two charged disks of radius 4.5 m separated by a distanc
ID: 2137534 • Letter: A
Question
A capacitor consists of two charged disks of radius 4.5 m separated by a distance s = 2 mm (see the figure). The magnitude of the charge on each disk is 34
A capacitor consists of two charged disks of radius 4.5 m separated by a distance s = 2 mm (see the figure). The magnitude of the charge on each disk is 34 mu C. Consider points A, B, C, and D inside the capacitor, as shown in the diagram. The distance s1 = 1.3 mm, and the distance s2 = 0.5 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 s (delta)l along this path? What is VB - VA? Next, calculate the potential difference VC - VB. What is (delta)l along this path? What is VC - VB? Finally, calculate the potential difference VA - VC. What s (delta)l along this path? What is VA - VC?Explanation / Answer
a) E =Q/A*epsilon
= 34 *10^-6/(3.14*4.5^2*8.85*10^-12) = 6.04 *10^4 N/C
(b)The potential difference between points B and A is
VB - VA= E * s1
Here,s1=1.3 mm = 1.3 * 10-3m
or VB - VA= 6.04 *10^4 * 1.3 *10-3
or VB - VA= 78.52 V ---------(1)
c)The potential difference between points C and Bis
VC - VB= E * s2
Here,s2= 0.5 mm = 0.5 *10-3m
VC - VB= 6.04 *10^4 * 0.5 * 10-3= 30.2 V----------(2)
d) Adding equations (1) and (2),we get
VC - VA= 78.2 +30.2 = 108.4 V
()The distance between points C and A is
?L = [(1.3)^2 +(0.5)^2]^1/2 = 1.39 mm = 1.39 * 10-3m
Therefore,we get
?Lcos? = 1.39 * 10-3 * cos(45o)= 1.39 * 10-3 * 0.7072 = 0.98 * 10-3 m
Therefore,the value of ?V along the diagonal path from Ato C is
VC - VA= E?Lcos?
or VC - VA= 6.04 *10^4 * 0.98* 10-3
or VC - VA=58.8 V
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