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ID: 1703576 • Letter: #
Question
<p><img src="https://s3.amazonaws.com/answer-board-image/b79df520-ea9d-44d9-a50f-98914c7db0a7.jpg" alt="image from custom entry tool" /></p><p>A parallel-plate capactior with spacing b and area A is connected to a battery of voltage V as shown above. Initially the space between the plates is empty. Make the following determinations in terms of the given symbols.</p>
<p>a. Determine the electric field between the plates</p>
<p>b. Determine the charge stored on each capacitor plate</p>
<p>A copper slab of thickness a is now inserted midway between the plates with a height of a.</p>
<p>c. Determine the electric field in the spaces above and below the slab.</p>
<p>d. Determine the ratio of capacitances C with copper/C original when the slab is inserted</p>
<p> </p>
Explanation / Answer
distance between parallel plates is d
area of plates is A
voltage applied to the capacitor is V
a) electric field between the plates is
E = V / b
b) charge stored on the each capacitor pate
C = Q / V
Q = C V
where C is capacitance of the Capacitor
C = _o A / b
therefore charge stored in the capacitor
Q = ( _o A / b)V
c) electric field in above and below the slab is
is E = q / k o A
d) capacitance of the capaciotr after the slab is inserted
C _Cu = _oA k / [kb - a(k -1 ) ]
where k is dielectric constant of slab
C _cu / C
= { _oA k / [kb - a(k -1 ) ] } / [_o A / b]
= k b / [kb - a(k -1 ) ]
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