Answer to Question #170512 in Physics for Elm

Question #170512

A parallel plate capacitor having a plate area of 0.70 m2 and a plate separation of 1.0 mm is connected to a source with a voltage of 50 V. Find the capacitance, the charge on the plates, and the energy of the capacitor b) when the capacitor contains polystyrene


1
Expert's answer
2021-03-10T17:15:54-0500

1. The capacitance of the parallel plate capacitor is given as follows:


C=ε0εSdC = \dfrac{\varepsilon_0\varepsilon S}{d}

where ε0=8.851012F/m\varepsilon_0 = 8.85 \cdot 10^{−12}F/m is the electric constant, ε=2.5\varepsilon = 2.5 is the permittivity of polystyrene, S=0.7m2S = 0.7m^2 is the area of the plates, and d=1mm=103md = 1mm = 10^{-3}m is the distance between the plates.

Thus, obtain:


C=8.8510122.50.710315.5×109FC = \dfrac{8.85 \cdot 10^{−12}\cdot 2.5\cdot 0.7}{10^{-3}} \approx 15.5\times10^{-9}F

2. The charge on the plates (on one plate) is:


q=CVq = CV

where V=50VV = 50V is the voltage of the source. Thus, obtain:


q=15.5×109F50V7.8×107Cq = 15.5\times 10^{-9}F\cdot 50V \approx 7.8\times 10^{-7}C

3. The energy stored in the capacitors is:


W=CV22W=15.5×109×50221.9×105JW = \dfrac{CV^2}{2}\\ W = \dfrac{15.5\times10^{-9}\times 50^2}{2}\approx 1.9\times 10^{-5}J

Answer. a) 15.5×109F15.5\times10^{-9}F, b) 7.8×107C7.8\times 10^{-7}C, c) 1.9×105J1.9\times 10^{-5}J.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment