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A slab of glass dielectric is inserted into a parallel plate capacitor
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[QUOTE="hidemi, post: 6457339, member: 686975"] [B]Homework Statement:[/B] A parallel-plate capacitor, with air dielectric, is charged by a battery, after which the battery is disconnected. A slab of glass dielectric is then slowly inserted between the plates. As it is being inserted: A. a force repels the glass out of the capacitor B. a force attracts the glass into the capacitor C. no force acts on the glass D. a net charge appears on the glass E. the glass makes the plates repel each other The answer is B [B]Relevant Equations:[/B] C = k(ε*Area)/distance = Q/V = Q/ (E*distance) F = QE I use the following equations to understand this question/answer. First, C = k(ε*Area)/distance = Q/V = Q/ (E*distance) As a slab of glass is added, k increases and thus E decreases. F=QE, as E decreases, force decreases as well. How does this relate to the 'force attracts the glass into the capacitor' Let me know if my thoughts/logic on this is right. In addition, is there a less physic-based explanation? Thanks. [/QUOTE]
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A slab of glass dielectric is inserted into a parallel plate capacitor
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