As part of a science fair project you want to construct several capacitors with different dielectrics inside. The conducting plates will be made from aluminum (with surface area A). The first capacitor uses a plane of glass with thickness d, surface area A, and dielectric constant κ (Figure 1). It has capacitance Cglass. The second uses a piece of plastic with thickness d/2, surface area A, and dielectric constant 1.5κ (Figure 2). It has capacitance Cplastic. The third one uses both glass and plastic, and is shown in Figure 3. This configuration has capacitance Ctotal. Part (a) Enter an expression for the theoretical capacitance, Cglass, of the setup using only glass (Figure 1) in terms of A, d, κ, and ε0. Part (b) Now enter an expression for the capacitance using only plastic, Cplastic, in terms of the capacitance using only glass, Cglass. Part (c) Now you construct the last capacitor, using both plastic and glass, with no gap between the glass and plastic. Enter an expression for the capacitance of this system, Ctotal, in terms of A, d, κ, and ε0. Part (d) Calculate the capacitance of the final system, Ctotal, in nanofarads, assuming A = 19.7 cm2, d = 0.37 mm, κ = 3.1.
Dielectric Constant Of Water
Water constitutes about 70% of earth. Some important distinguishing properties of water are high molar concentration, small dissociation constant and high dielectric constant.
Electrostatic Potential and Capacitance
An electrostatic force is a force caused by stationary electric charges /fields. The electrostatic force is caused by the transfer of electrons in conducting materials. Coulomb’s law determines the amount of force between two stationary, charged particles. The electric force is the force which acts between two stationary charges. It is also called Coulomb force.
Problem 4: As part of a science fair project you want to construct several capacitors with different dielectrics inside. The
Part (a) Enter an expression for the theoretical capacitance, Cglass, of the setup using only glass (Figure 1) in terms of A, d, κ, and ε0.
Part (b) Now enter an expression for the capacitance using only plastic, Cplastic, in terms of the capacitance using only glass, Cglass.
Part (c) Now you construct the last capacitor, using both plastic and glass, with no gap between the glass and plastic. Enter an expression for the capacitance of this system, Ctotal, in terms of A, d, κ, and ε0.
Part (d) Calculate the capacitance of the final system, Ctotal, in nanofarads, assuming A = 19.7 cm2, d = 0.37 mm, κ = 3.1.
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