Two parallel rectangular plates measuring 80 mm by 12 cm are separated by 4 mm of mica and carry an electric charge of 0.48 MC. The voltage between the plates is [tex]$500 V$[/tex]. Calculate:
(a) The electric flux density.
(b) The electric field strength.
(c) The capacitance of the capacitor, in picofarads, if the relative permittivity of mica is 5.
Answer (2)
Calculate the area of the plates: A = 0.08 × 0.12 = 0.0096 m 2 .
Calculate the electric flux density: D = A Q = 0.0096 0.48 × 1 0 6 = 50 × 1 0 6 C/m 2 .
Calculate the electric field strength: E = d V = 0.004 500 = 125000 V/m .
Calculate the capacitance: C = d ϵ r ϵ 0 A = 0.004 5 × 8.854 × 1 0 − 12 × 0.0096 ≈ 106 pF .
The final answers are: Electric flux density is 50 MC/m 2 , electric field strength is 125 kV/m , and capacitance is 106 pF .
Explanation
Problem Setup and Given Data We are given two parallel rectangular plates with specific dimensions, separation, charge, voltage, and relative permittivity. We need to calculate the electric flux density, electric field strength, and capacitance.
Calculating the Area The dimensions of the rectangular plates are 80 mm by 12 cm. Converting these to meters, we have 0.08 m and 0.12 m. The area A is then calculated as: A = 0.08 m × 0.12 m = 0.0096 m 2
Stating Other Given Values The separation distance between the plates is 4 mm, which is 0.004 m. The charge on the plates is 0.48 MC, which is 0.48 × 1 0 6 C. The voltage between the plates is 500 V. The relative permittivity of mica is 5.
Calculating Electric Flux Density (a) The electric flux density (D) is calculated using the formula: D = A Q = 0.0096 m 2 0.48 × 1 0 6 C = 50 × 1 0 6 C/m 2 = 50 MC/m 2
Calculating Electric Field Strength (b) The electric field strength (E) is calculated using the formula: E = d V = 0.004 m 500 V = 125000 V/m = 125 kV/m
Calculating Capacitance (c) The capacitance (C) is calculated using the formula: C = d ϵ r ϵ 0 A where ϵ 0 is the permittivity of free space ( 8.854 × 1 0 − 12 F/m). C = 0.004 m 5 × 8.854 × 1 0 − 12 F/m × 0.0096 m 2 = 1.06248 × 1 0 − 10 F Converting to picofarads: C = 1.06248 × 1 0 − 10 F × 1 0 12 pF/F = 106.248 pF Rounding to three significant figures, we get 106 pF.
Final Answer Therefore, the electric flux density is 50 MC/m 2 , the electric field strength is 125 kV/m , and the capacitance is approximately 106 pF .
Examples
Understanding capacitance is crucial in designing electronic circuits. For example, in a smartphone, capacitors store electrical energy, allowing the device to function smoothly. The principles used to calculate capacitance, electric field strength, and electric flux density in this problem are directly applicable to designing and optimizing the performance of such components in various electronic devices, ensuring efficient energy storage and release.
The electric flux density is 50 MC/m², the electric field strength is 125 kV/m, and the capacitance of the capacitor is approximately 106 pF.
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