The plates of a parallel-plate capacitor have a separation of 5.82 mm , and each has an area of 14.8 cm². Each plate carries a charge of magnitude 5.10x10-8 C. The plates are in vacuum. You may want to review (Pages 575 - 578) . For related problemsolving tips and strategies, you may want to view a Video Tutor Solution of Properties of a parallel-plate сараcitor.
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I got this question wrong and keep getting different values with the calculations I am doing. If possible, please include the steps so I may understand where I went wrong with my calculations.
(a). What is the capacitance? (in F)
(b). What is the potential difference between the plates? (in v)
(c). What is the magnitude of the electric field between the plates? (in V/m)
d = 5.82 mm
A= 14.8 cm2
Q= 5.1× 10-8 C
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