Two 2.0 cm multiplied by 2.0 cm plates that form a parallel-plate capacitor are

Question

Two 2.0 cm multiplied by 2.0 cm plates that form a parallel-plate capacitor are charged to ±0.775 nC. What are the electric field strength inside and the potential difference across the capacitor if the spacing between the plates is (a) 1.2 mm and (b) 2.4 mm.

ELECTRIC FIELD STRENGTH (N/C) POTENTIAL DIFFERENCE (V)
(a) __________________________ (a) 262.71 V
(b) __________________________ (b) 525.52 V

I have already solved for the Potential Difference (see above). Please solve for the Electric Field Strength in N/C.

Explanation / Answer

The relation between the electri field and potential difference is              V = Ed a) The electric field strength is             E = V/d                 = ( 262.71 V)/(1.2x 10^-3 m)                 = 2.19x10^+5 N/C b) The electric field strength is              E = V/d                 = ( 525.52 V)/(2.4x 10^-3 m)                 = 2.19x10^+5 N/C b) The electric field strength is              E = V/d                 = ( 525.52 V)/(2.4x 10^-3 m)                 = 2.19x10^+5 N/C                 = ( 525.52 V)/(2.4x 10^-3 m)                 = 2.19x10^+5 N/C

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