Question

Part A An electric field of 9.00x105 V/m is desired between two parallel plates, each of area 45.0 cm and separated by 2.45mm of air. What charge must be on each plate? 2 Express your answer using three significant figures and include the appropriate units. QValue Units

Answer 1

$\text{Solution :}\\ \text{Given :}\\ *\text{Area of plates}\ \ A= 45.0\ cm^2= 45.0\times10^{-4}m^2\\ *\text{Distance between plates }\ \ \ d = 2.45\ mm = 2.45\times 10^{-3}m\\ *\text{Vacuum permittivity}\ \ \epsilon_o = 8.85\times 10^{-12}F/m\\ \ \\ \text{The capacitance of the capacitor is given by;}\\ C= \dfrac{\epsilon_o \ A}{d}\\ C= \dfrac{(8.85\times 10^{-12}F/m)\times (45.0\times10^{-4}m^2)}{2.45\times 10^{-3}m}\\ C=1.62\times 10^{-11} \ F\\ \ \\ Now;\\ \text{Potential difference between plates:}\\ V= E.d\\ V= (9.00\times10^5V/m)\times(2.45\times 10^{-3}m)\\ V= 2205\ V \ \\ \ \\ \text{Charges on each plate} \ :\\ Q= C\ V\\ Q= (1.62\times 10^{-11} \ F)\times 2205\ V\\ Q= 35.72\times 10^{-9}C\\ Q= 35.72\ nC$