A capacitor has a capacitance of 55 pF. In the charging process electrons are removed from one ate and placed on the other plate. When the potential difference between the plates is 5.3 v how many electrons have been transferred?
C=q/U =Ne/U ,
N=CU/e=55•10⁻ ¹²•5.3/1.6•10⁻ ¹⁹ =1.82•10⁹
To determine the number of electrons transferred, we need to first understand the relationship between capacitance, charge, and potential difference.
The formula relating these variables is:
Q = C * V,
where Q is the charge, C is the capacitance, and V is the potential difference.
Given that the capacitance is 55 pF (picofarads) and the potential difference is 5.3 V, we can substitute these values into the formula:
Q = 55 pF * 5.3 V.
To proceed with the calculation, we need to convert the capacitance to farads, as the formula requires the value in this unit.
Remember that 1 pF = 1 × 10^(-12) F (farads). Hence:
C = 55 pF = 55 × 10^(-12) F.
Now we can substitute the capacitance value into the formula:
Q = (55 × 10^(-12) F) * 5.3 V.
Multiply the capacitance and the potential difference:
Q = 2.915 × 10^(-10) C.
The charge represents the total transferred electrons, and since we know that 1 electron has a charge of approximately 1.6 × 10^(-19) C, we can calculate the number of electrons transferred by dividing the charge by the elementary charge:
Number of electrons transferred = Q / (1.6 × 10^(-19) C).
Substituting in the value for Q:
Number of electrons transferred = (2.915 × 10^(-10) C) / (1.6 × 10^(-19) C).
Performing the division:
Number of electrons transferred ≈ 1.822 × 10^9 electrons.
Therefore, approximately 1.822 billion electrons have been transferred during the charging process.