The voltage (in kv) must be applied to an 8.90 nf capacitor to store 0.175 mc of charge is
19662.921V
Having knowledge of the relationship between the charge Q on a capacitor with an applied voltage V and the capacitance C
Q=CV
Therefore, we can add to the previous equation to use the capacitance value and the total charge on the capacitor's plates to calculate the applied voltage. Consequently, we have
[tex]V=\frac{Q}{C} \\\\V= \frac{0.175 x 10^{-3} }{8.90 x 10^{-9} } \\\\\\V=19662.921 V[/tex]
Therefore the required voltage is 19662.921 V
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