Question

An engineering students designs and builds a capacitor from some scrap iron. One piece of iron is a square with side lengths of 10 cm, while the other piece is a rectangle that is 20 cm long and 10 cm wide. The student knows they can keep the plates separated by a distance of 1.2 micrometers.

a. What is the capacitance of this setup?

b. If the capacitor is fully charged using a 1.5 V battery, how much charge is built up on the plates?

c. How much energy is stored?

d. After performing the experiment, the student inserts a dielectric with a k value of 1.2. What values would this change?

Answer 1

PART (a):

Area of square plate, A₁ = (0.1 m)² = 0.01 m²

Area of rectangular plate, A₂ = (0.1 cm) x (0.2 cm) = 0.02 m²

Plate separation, d = 1.2 μm = 1.2 x 10^-6 m

Therefore, Capacitance of this setup is given by:

OR

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PART (b):

Potential, V = 1.5 V

Therefore, Charge in capacitor is given by:

OR

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PART (c):

Energy (U) stored in capacitor is given by:

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PART (d):

Inserting a dielectric inside the capacitor would change the value of:

• Capacitance (C' = 1.2 x C) = 118 nF

Since the capacitance changes, this will change the values of :

• Charge (Q' = 1.2 Q) = 177 nC
• Energy stored (U' = 1.2 U) = 1.3275 x 10^-7 J

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