Two parallel plates each of area A are spaced adistance d and connected by wires

Question

Two parallel plates each of area A are spaced adistance d and connected by wires to the terminals of abattery voltage . (a) Find an expression for the charge on the positive plate,the electric field strength, and the potential difference betweenthe two plates in magnitude. (b) The wires are disconnected and insulated handles are usedto pull the plates apart to a new spacing of 4d. Now find thecharge on the positive plate, the electric field strength, and thepotential difference between them in magnitude. Two parallel plates each of area A are spaced adistance d and connected by wires to the terminals of abattery voltage . (a) Find an expression for the charge on the positive plate,the electric field strength, and the potential difference betweenthe two plates in magnitude. (b) The wires are disconnected and insulated handles are usedto pull the plates apart to a new spacing of 4d. Now find thecharge on the positive plate, the electric field strength, and thepotential difference between them in magnitude.

Explanation / Answer

a) Let A be the area of the two parallel plates of the capacitorand d be the seperation between the plates of thecapacitor. And be the electric potential difference between theplates of the capacitor. Now formula for calculating the value of the capacitance ofthe capacitor is                    C = Ao / d Now formula for calculating the vlaue of the charge on thepositive plate of the capacitor is              q = C                 = (Ao / d) Formula for calculating the value of the electric fieldstrength between the plates of the capacitor is             E = / d b) If the seperation between the plates is increased to4d then the charge on the positive plate remainsconstant. And the potenital difference is increased by 4 times.

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