Two 2.60 cm × 2.60 cm plates that form a parallel-plate capacitor are charged to

Question

Two 2.60 cm × 2.60 cm plates that form a parallel-plate capacitor are charged to ± 0.708 nC .

Part A. What is the electric field strength inside the capacitor if the spacing between the plates is 1.50 mm ?

Part B. What is potential difference across the capacitor if the spacing between the plates is 1.50 mm ?   

Part C. What is the electric field strength inside the capacitor if the spacing between the plates is 3.00 mm ?   

Part D. What is the potential difference across the capacitor if the spacing between the plates is 3.00 mm ?   

Explanation / Answer

Area=6.76*10^(-4)

capacitance=C=epsilon0*Area/distance=3.99*10^(-12) F

voltage=Charge/Capacitance=177.43 V

electric field=voltage/distance=118286.6667 N/C

if distance=3 mm

C will be halved

so Voltage will doubled=354.86 V

Electric field=voltage /distance=118286.6667

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