Problem 21.13 - Enhanced with Feedback What is the electric field strength insid

Question

Problem 21.13 - Enhanced with Feedback What is the electric field strength inside the capacitor if the spacing between the plates is 1.00 mm? Express your answer with the app ropriate units. Two 2.70 cm × 2.70 cm plates that form a parallel-plate capacitor are charged to ± 0.708 nC You may want to review ( pages 678-683) E-Value Units Submit My Answers Give Up Incorrect; One attempt remaining; Try Again Part B What is potential difference across the capacitor if the spacing between the plates is 1.00 mm? Express your answer with the app ropriate units. V= 110 V Submit My Answers Give Up Correct Here we learn how to calculate the potential difference between the plates of a capacitor Part C What is the electric field strength inside the capacitor if the spacing between the plates is 2.00 mm? Express your answer with the appropriate units. E-Value Units Submit My Answers Give Up

Explanation / Answer

Given parallel plate capacitor with area of the plates is A = 2.7*2.7 cm^2 = 0.000729 m^2

charge q = 0.708 n C

we know that teh capacitance of the parallel plate capacitor is C = epsilon not *A/d and

electric field is E = sigma /epsilon not = V/d

where sigma is the surface charge density sigma = q/A = 0.708*10^-9/(0.000729) C/m^2 = 9.7119342*10^-7 C/m^2

E = (9.7119342*10^-7)/(8.854*10^-12) = 109689.792 N/C

OR

and the relation between q,C,V is q = C*V ==> V = q/C = q/(epsilon not *A/d) = 0.708*10^-9/(8.854*10^-12*0.000729/0.001) V = 109.689792 V

PartA

now the electric field when the separation is 1 mm , E = V/d = 109.689792/(0.001) N/C = 109689.792 N/C

Part C

q = C*V ==> V = q/C = q/(epsilon not *A/d) = 0.708*10^-9/(8.854*10^-12*0.000729/0.002) V = 219.37958338 V

now the electric field when the separation is 2 mm , E = V/d = 219.37958338/(0.002) N/C = 109689.792 N/C

Part D

v when d= 2 mm is  

V = E*d = 109689.792 *0.002 V = 219.379584 V

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