Respuesta :
Answer:
0.0002 C.
Explanation:
Charge: This can be defined as the ratio of current to time flowing in a circuit. The S.I unit of charge is Coulombs (C)
Mathematically, charge can be expressed as
Q = CV ................................. Equation 1
Where Q = amount of charge, C = capacitance of the capacitor, V = potential difference across the plates.
Given: C = 2.0-μF = 2×10⁻⁶ F, V = 100 V.
Substitute into equation 1
Q = 2×10⁻⁶× 100
Q = 2×10⁻⁴ C
Q = 0.0002 C.
The amount of charge accumulated = 0.0002 C
Answer:
The capacitor accumulates 0.2 mC
Explanation:
The capacitance (with units Faraday) of a capacitor with a charge Q (with units Columbus) and a potential difference ΔV (with units Volts) is:
[tex]C=\frac{Q}{\varDelta V} [/tex] (1)
In our case C=[tex]2.0\times10^{-6}F [/tex] and ΔV=[tex] 100V[/tex], so we can solve (1) for Q and use those values:
[tex] C=\frac{Q}{\varDelta V}[/tex]
[tex] Q=C{\varDelta V}=(2.0\times10^{-6}F)(100V)[/tex]
[tex] Q=2.0\times10^{-4}C[/tex]