two resistors of 20 ohm each are connected in a parallel with a battery of 10V. The total current passing through circuit is
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Answer:
The number of electrons that must be moved from one electrode to the other to accomplish this is 1.4 X 10⁹ electrons.
Explanation:
<u>Step 1:</u> calculate the charge on each electrode
Given;
Electric field strength = 2.0 X 10⁶ N/C
The distance between the electrode = 1mm = 1 X 10⁻³ m
Electric field strength (E) = Force (F)/Charge (q)
where;
E is the electric field strength = 2.0 X 10⁶ N/C
K is coulomb's constant = 8.99 X 10⁹ Nm²/C²
r is the distance between the electrodes = 1 X 10⁻³ m
q is the charge in each electrode = ?
= 0.2225 X 10⁻⁹ C
The charge on each electrode is 0.2225 X 10⁻⁹ C
<u>Step 2:</u> calculate the number of electrons to be moved from one electrode to the other.
1 electron contains 1.602 X 10⁻¹⁹ C
So, 0.2225 X 10⁻⁹ C will contain how many electrons ?
= (0.2225 X 10⁻⁹)/(1.602 X 10⁻¹⁹)
= 1.4 X 10⁹ electrons
Therefore, the number of electrons that must be moved from one electrode to the other to accomplish this is 1.4 X 10⁹ electrons.