Basic Electric Science Questions and Solutions
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A number of questions pertaining to parallel plate capacitors are presented and solutions showing how to determine the Electric Field strength, Charge per unit area and other parameters such as charged stored are presented.
The second part of the exercise shows how to determine which bulbs light in a series parallel circuit with switches when variuos switches are opened and closed
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Solution Summary
A number of questions pertaining to parallel plate capacitors are presented and solutions showing how to determine the Electric Field strength, Charge per unit area and other parameters such as charged stored are presented.
The second part of the exercise shows how to determine which bulbs light in a series parallel circuit with switches when variuos switches are opened and closed
Solution Preview
See attached files.
QUESTION1
PART A: Solution.
Here use Electric field (E) is defined as voltage potential across the plates per unit distance between plates
E = V/d {need to convert d from mm into SI units of length ie metres)
Now d = 5 mm = 0.005 m so
E = 20/0.005 = 4000 V/m
PART B: Solution.
For the second part use Gauss relationship that the Electric field is equal to the Charge per unit area density (sigma) divided by the dielectric constant (epsilon)
For air dielectric Epsilon = Epsilon0 = 8.854 x 10^-12 F/m
Hence use E = sigma/epsilon and rearrange for sigma = epsilon0*E
Charge desnisty Sigma = 8.854 x 10^-12 x 4000 = 35.4 nC/m2
PART C: Solution.
Here I have used the fact that the energy stored in a capacitor is given by
E = 0.5*C*V^2
Also Capacitance is given by C = Epsilon0*A/d
Note is important to make sure you are using SI units throughout so convert
d1 = 5mm to meters so d = 0.005 m in first instance, d2 = 10mm = 0.01 m in second instance
Area of plates A = 1 cm2 = 10 x 10^-4 m2
{again a ...
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