A parallel-plate capacitor carries a constantcharge Q (i.e., + Q on one plate, a

Question

A parallel-plate capacitor carries a constantcharge Q (i.e., +Q on one plate, and -Qon the other). When a dielectric is inserted between the plates,what happens to the voltage difference between the plates, and thepotential energy stored in the capacitor?

(a) The voltage increases, and the potential energy decreases.
(b) The voltage and potential energy both increase.
(c) The voltage and potential energy both decrease.
(d) The voltage decreases, and the potential energy increases.
(e) None of the above.

I picked (c) for this one because if the voltage decreases thenaccording to C = (Q/V) the capacitance and charge increase and ifthe charge increases the potential energy increases.Right?

A parallel-plate capacitor is attached to abattery that maintains a constant voltage differenceV across the plates. When a dielectric is insertedbetween the plates, what happens to the charge on the plates, andthe potential energy stored in the capacitor?

(a) The charge decreases, and the potential energy increases.
(b) The charge and potential energy both increase.
(c) The charge increases, and the potential energy decreases.
(d) The charge and potential energy both decrease.
(e) None of the above.

I picked (b) for this one but I wasn't sure because can't it goeither way? Like they could either both increase or they could bothdecrease? How do you know if it increases or decreases?
Will rate lifesaver! Thanks!

Explanation / Answer

(a). Charge remains constant . After insertion of dielectric capacitance increases by Ktimes where K = dielectric constant we know C = Q / V               Q = CV Capacitance C is increases by K times, But charge Q remainssmae this is possible only when Voltage decreases by K times So, Voltage decreases. i.e., new capacitance C ' = KC new voltage V ' = V / K Initial potential energy U = ( 1/ 2) C V ^ 2 So, New potential energy U ' = ( 1/ 2) C ' V ' ^ 2                                             = ( 1/ 2) KC * ( V / K ) ^ 2                                            = U / K i.e., potential decreases Therefore option ( c ) is correct (b). Voltage remains constant . we know Q = CV               Q C SInce voltage isconstant      ----( 1) After insertion of dielectric capcitance increases and itsvalue C ' = KC But from eq ( 1 ) , charge Q is directly proportional tocapacitance C value So, charge is also increases. and new charge Q ' = K Q we know Initial potential energy U = Q ^ 2 /2C new potential energy U ' = Q ' ^ 2 / 2C ,                                      = ( K Q ) ^ 2 / [ 2KC ]                                      = K U i.e., potential energy increases. Therefore option ( d ) is correct                                        

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