how many 0.25-pF capacitors must be connected in parallel in order to store 1.2uC of charge when connected to a battery providing a potential difference of 10v?
C = q/V
so
q = C V
so each stores 2.5 * 10^-12 C
so we need 1.2*10^-6/2.5*10^-12 = .48*10^6
=480,000
1.2x10^-6 / 0.25x10^-12 = 4.8x10^6
but C times ten volts = q per capacitor
To find the number of capacitors required to store a certain amount of charge, we can use the formula:
Q = C * V
Where:
Q = charge (in coulombs)
C = capacitance (in farads)
V = voltage (in volts)
We have the following information:
Q = 1.2 μC (microcoulombs) = 1.2 * 10^-6 C
V = 10 V
C = 0.25 pF (picofarads) = 0.25 * 10^-12 F
Let's plug these values into the formula and solve for C:
1.2 * 10^-6 C = (0.25 * 10^-12 F) * 10 V
Simplify the equation:
1.2 * 10^-6 C = 2.5 * 10^-12 F * 10 V
Now, rearrange the equation to solve for the number of capacitors (N):
N = Q / (C * V)
Substitute the given values:
N = (1.2 * 10^-6 C) / ((0.25 * 10^-12 F) * 10 V)
Simplify:
N = 1.2 * 10^-6 C / (2.5 * 10^-12 F * 10 V)
N = (1.2 / 2.5) * (10^-6 / (10^-12 * 10))
N = 0.48 * 10^6 / 10^-11
N = 4.8 * 10^7 / 10^-11
N = 4.8 * 10^7 * 10^11
N = 4.8 * 10^18
Therefore, you would need approximately 4.8 x 10^18 0.25-pF capacitors connected in parallel to store 1.2 μC of charge when connected to a 10 V battery.