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3 years ago

A 0.55-μF capacitor is connected to a 3.5-V battery. How much charge is on each plate of the capacitor?

Physics

2 answers:

yan [13]3 years ago

6 0

Answer:

1.925 μC

Explanation:

Charge: This can be defined as the product of the capacitance of a capacitor and the voltage. The S.I unit of charge is Coulombs (C)

The formula for the charge stored in a capacitor is given as,

Q = CV ................... Equation 1

Where Q = charge, C = Capacitor, V = Voltage.

Note: 1 μF = 10⁻⁶ F

Given: C = 0.55 μF = 0.55×10⁻⁶ F, V = 3.5 V.

Substitute into equation 1

Q = 0.55×10⁻⁶×3.5

Q = 1.925×10⁻⁶ C.

Q = 1.925 μC

Hence the charge on the plate = 1.925 μC

andrew11 [14]3 years ago

4 0

Answer:

1.925μC

Explanation:

The relationship between charge (Q), voltage/potential difference (V) and capacitance (C) of a parallel plate capacitor is given by;

Q = CV ----------------------(i)

Where, according to the question;

C = 0.55 μF = 0.55 x $10^{-6}$ F

V = 3.5V

Substitute the values of C and V into equation (i);

=> Q = CV

=> Q = 0.55 x $10^{-6}$ x 3.5

=> Q = 1.925 x $10^{-6}$ C

=> Q = 1.925μC

Therefore the amount of charge on each plate of the capacitor is 1.925μC

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