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Question: An \(8\;{{\mu F}}\) Capacitor is connected across a \(220\;{\text{V}}\), \(50\;{\text{Hz}}\) line. W...

An 8  μF8\;{{\mu F}} Capacitor is connected across a 220  V220\;{\text{V}}, 50  Hz50\;{\text{Hz}} line. What is the peak value of charge through the capacitor?
A. 2.5×10−3  C2.5 \times {10^{ - 3}}\;{\text{C}}
B. 2.5×10−4  C2.5 \times {10^{ - 4}}\;{\text{C}}
C. 5×10−5  C5 \times {10^{ - 5}}\;{\text{C}}
D. 7.5×10−2  C7.5 \times {10^{ - 2}}\;{\text{C}}

Explanation

Solution

Amount of charge stored in the conductor is proportional to the voltage and the constant of proportionality is Capacitance. The capacitance is the ratio of the charge on the conducting plates to the voltage across them.

Formula used:
The charge on the capacitor Q=CΔVQ = C\Delta V
Where CC is the capacitance and ΔV\Delta V is the voltage drop across the capacitor.

Complete step by step solution:
Understand that, Maximum voltage through the capacitor is,
Vmax⁡=2V{V_{\max }} = \sqrt 2 V
Substitute 1212 V for V
Vmax⁡=2(12V) =122V  {V_{\max }} = \sqrt 2 (12{\text{V}}) \\\ = 12\sqrt 2 {\text{V}} \\\
Write down the expression for the peak value of the charge through the capacitor is given as,
Q=CVmax⁡Q = C{V_{\max }}
Substitute 8×10−6  F8 \times {10^{ - 6}}\;{\text{F}} for CC and 2202  V220\sqrt 2 \;{\text{V}} for Vmax⁡{V_{\max }} values in above expression.
Q=8×10−6  F×2202  V  = 2.5×10−3  C  Q = 8 \times {10^{ - 6}}\;{\text{F}} \times 220\sqrt 2 \;{\text{V}} \\\ {\text{ = 2}}{\text{.5}} \times {\text{1}}{{\text{0}}^{ - 3}}\;{\text{C}} \\\
The peak value of charge through the capacitor is 2.5×10−3  C{\text{2}}{\text{.5}} \times {\text{1}}{{\text{0}}^{ - 3}}\;{\text{C}}.

∴\therefore Hence, the option A is correct.

Additional information:
The energy is stored in between the negative charge plate and the positively charged plates while charging. The charged-up capacitor will store the potential energy. Capacitance means the charge on the conductor which causes the voltage across it. The voltage drop will be greater for the more charged capacitor. An electric field is formed between the capacitor plates when the positive charge on one plate and negative charge on one plate is accumulated. This happens when the voltage is applied across the terminals of the capacitor. The attraction of the opposite charges in the plates causes the storage of energy in the capacitor. The charge is flowing through the source. The plate area of the capacitor and the dielectric used can determine the amount of energy the capacitor can store. The maximum voltage across the capacitor is 2\sqrt 2 times the voltage applied across its terminals. The unit of capacitance is measured in Farad.

Note:
The charge is always the product of capacitance and the voltage. When the voltage increases the charge passing through the capacitor will always increase. The peak value of the charge is the maximum charge attained by the capacitor in one cycle. The RMS value of voltage means the square root of the mean of squares of the instantaneous values of voltage.