A capacitor is completely charged with 560 nC by a voltage source that has 325 V

Question

A capacitor is completely charged with 560 nC by a voltage source that has 325 V What is its capacitance? Tries 0/2 Submit Answer Now the plates of the charged capacitor are pushed together with the voltage source still connected The voltage drop between the plates decreases. The energy stored in the capacitor remains the same. The voltage drop between the plates increases. The charge on the plates increases. None of the above. Submit Answer Tries 0/2 The initial air gap of the capacitor above was 8 mm. What is the stored energy if the air gap is now 5 mm? Submit Answer Tries 0/2

Explanation / Answer

(a)

C= Q/V = 560 * 10^-9 C/ 325 V = 1.72 * 10^-9 F

(b)

Because the capacitor is still connected to the power supply the potential difference can't change. Moving the plates pushed together increases the capacitance, also increasing the charge stored by the capacitor.  

the charge stored on the plates increases is the correct option

(c)

U = 1/2 * CV^2 = 1/2 ( eo A/ d) V^2

U1 = QV/2 = 560* 10^-9) ( 325)/2= 9.1* 10^-5 J

U is proportional to d

so U2/ U1 = d2/ d1

U2 = U1 ( d2/ d1) = 9.1* 10^-5 ( 5/8) = 5.68 * 10^-5 J

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