Question

A capacitor consists of two closely spaced metal conductors of large area, separated by a thin insulating film. One of the many practical uses of a capacitor is to store up some energy, to be released in a sudden burst (like lighting up the flash bulb in a camera!) A decent (not too expensive) capacitor in such a camera might be 100 uF, and could release a total electrical energy of 4.8 J in a flash. What voltage difference would you need across this capacitor? How much charge is stored on the NEGATIVE PLATE of the capacitor in the previous question? If the two plates of the capacitor in the previous question have their separation increased by a factor of 8 while the charge on the plates remains constant, by what factor is the energy stored in the capacitor increased?

Answer 1

Solution a) The voltage difference between the capacitors are, The energy stored is given by, Here, E is the energy, C is the capacitance and V is the voltage. Rearrange the above term for Vas, 2E 2x4.8 100x10-5F 309.84V b) The charge stored on the negative plate of the capacitor is, The charge stored is, (100x10 F (309.84V)

c) When the separation is increased by a factor of 8 while the charge on the plates remains constant, the energy stored in the capacitor is increased by factor found as, We know that the expression for the capacitance of a parallel plate capacitor with dielectric constant k is, Here, So, when the separation between the plates get increased by a factor 8, then the energy stored get increased by a factor 1/8 or we can say that the energy get decreased by a factor