how many electrons would have to be removed from a coin to leave it with charge pf 10^-7c
Answers
Answered by
85
Q = ne
Here
- Q = charge
- n = number of electrons
- e = charge of 1 electron
Charge of 1 electron = 1.6 × 10^-19 Coulomb
Number of electrons = 10^-7 Coulomb / (1.6 × 10^-19 Coulomb)
= 625 × 10^9 electrons
= 625 billion electrons
625 billion electrons should be removed from [neutral] coin.
Here
- Q = charge
- n = number of electrons
- e = charge of 1 electron
Charge of 1 electron = 1.6 × 10^-19 Coulomb
Number of electrons = 10^-7 Coulomb / (1.6 × 10^-19 Coulomb)
= 625 × 10^9 electrons
= 625 billion electrons
625 billion electrons should be removed from [neutral] coin.
Answered by
5
Answer:
Explanation:
given data:
= 10^-7 Coulomb
formula used is :
Q = ne
we know
Charge of 1 electron = 1.6 × 10^-19 Coulomb
so,
Number of electrons
= 10^-7 Coulomb / (1.6 × 10^-19 Coulomb)
the ans will be:
= 625 × 10^9 electrons
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