The plates of a parallel plate capacitor are separated by 05...

The plates of a parallel plate capacitor are separated by 0.5 mm and the area of each plate is 400 cm^2. Determine the capacitance, if the dielectric is air. The charge stored in the capacitor when connected to a potential difference of 470 volts. The energy or work needed to move a single positive charge from high to low potential across the plates of the capacitor. Indicate the direction and value of the electric field due to the 470 V. (ε0 = 8.85 x 10^-12 C^2/N·m^2)

Transcript

00:01 Hello everyone, we are going to understand this question here given in the question given plates of parallel plate capacitor are separated by 0 .5 mm so separation between separation between separation between plates of parallel plate capacitor separation between plates of parallel plate capacitor that is represented by d which is 0 .5 millimeter or we can write 0 .5 in 10 to the core minus 3 meter and area of each plate area of each plate that is represented by 8 which is 400 centimeters square or we can write 400 in 10 to the 4 meters square because 1 meter is equal to 100 cm and we have to determine the capacitance of capacitor here directly constant of air is 1 so capacitance of capacitor capacitance of capacitor c is equal to epsilon naught a upon d here value of epsilon not is 8 .85 into 10 to the power minus 2n.

02:00 Value of k that is 400 into 10 to the or minus 4 upon separation between the plates that is 0 .5 into 10 to the power minus 3.

02:13 So after doing further calculation so we will we will get capacitors of capacitors c is equal to after calculation we will get 7 ,080 into 10 to the power minus 13.

02:32 So after calculation we will get 7 .08 into 10 to the power minus 10 ferret or we can write c is equal to 7 .08 7 .0 sorry here we can write 7 .08 and a strong ferret or we can write c is equal to 708 pico ferret now in the part v we have to find the charge stored in the capacitor the flight potential is 470 applied potential, b is equal to 470 volt, then charge stored in the capacitor, charge stored in capacitor that is represented by q which is equal to c into v.

03:39 Value of c that we have already calculated, 400 ,708 into 10 to be minus 12.

03:47 Value of c that is 470, c, so after calculation we will get.

03:57 332760 into 10 to the 1 .12 column so we can write 332 .76 nanog...

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