A capacitor has a capacitance of 3 pF. What potential difference would be required to store a charge of 21 pC?
Choose one answer.
a. 3 V
b. 21 V
c. 7 V
d. 0.14 V
C=q/V
V=q/C=21E-9/3E-9=7volts
To answer this question, we can use the formula for the capacitance of a capacitor:
C = Q / V
where C is the capacitance, Q is the charge stored in the capacitor, and V is the potential difference across the capacitor.
In this case, we have the capacitance C = 3 pF (picofarads) and the charge Q = 21 pC (picocoulombs). We can substitute these values into the formula and solve for V:
3 pF = 21 pC / V
To find V, we can multiply both sides of the equation by V and then divide both sides by 3 pF:
V = (21 pC / 3 pF)
It is important to note that pF (picofarads) and pC (picocoulombs) are both units in the metric system, so we can cancel them out in the division:
V = (21 C / 3 F)
Simplifying the equation further:
V = 7 V
So, the potential difference required to store a charge of 21 pC in a capacitor with a capacitance of 3 pF is 7 volts.
Therefore, the correct answer is c. 7 V.