1.) A parallel plate capacitor of plate area 115cm^2 and a plate separation of 1
ID: 1976844 • Letter: 1
Question
1.) A parallel plate capacitor of plate area 115cm^2 and a plate separation of 1.23cm. A potential difference of 86.6 Volts is applied between the plates. Calculatea.) the capacitance of the capacitor if air is filling the gap between the plates
b.) the free charge onthe plates
c.) the electric field between the plates
d.) the electric potential energy stored in the capacitor
e.) the capacitance of the capacitor if a dielectric medium of dielectric constant 2.61 is filling the gap between the plates.
2.) A diffraction grating had 12600 rulings spaced over the width 0.026m. It is illuminated by a doium doublet of wavelengths 589nm and 589.95nm. Calculate
a.) the slit grating spacing
b.) angle at which the first order maximum occur for the light of wavelength 589nm
c.) the resolving power of the grating
Explanation / Answer
a similar question is solved below, but with different values. hope this helps you. 1) 1. The capacitance 'C0' before the di-electric slab is inserted is (1 mark) 2. The charge on the plates is given by Q = C0V = (8.2110-12 )(85.5) = 7.0210-10 C (1 mark) The battery is disconnected and then the di-electric slab is introduced. So, the charge on the plates of the capacitor does not change. 3. The electric field in the region between the plates and the di-electric slab Where, is the charge density , Where 'Q' is the charge on the plates of a capacitor of area 'A' = 6900 Vm-1 (1 mark) 4. The electric field in the di-electric slab is = 2.64 103 Vm-1 (1 mark) 5. The potential difference between the plates is V = Eo (1.24 - 0.78) 10-2 - E1(0.78) 10-4 Substituting the values of 'E0' and 'E1', we get V 52.3 V (1 mark)
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