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One million electrons are added to a rod. The total charge on the rod is 1) 10⁻¹³ C 2) -1.6 x 10⁻¹³ C 3) +1.6 x 10⁻¹² C 4) 10⁻¹² C

Concept: Quantization of charge

Question

One million electrons are added to a rod. The total charge on the rod is
1) 10⁻¹³ C
2) -1.6 x 10⁻¹³ C
3) +1.6 x 10⁻¹² C
4) 10⁻¹² C

Try it yourself first

Hint: Recall the formula that relates the total charge to the number of electrons and the charge of a single electron.

Step-by-step solution

  1. 1

    Given Data

    Number of electrons added, n=1 million=106n = 1 \text{ million} = 10^6
  2. 2

    Charge of Electron

    Charge of one electron, e=−1.6×10−19 Ce = -1.6 \times 10^{-19} \text{ C}
  3. 3

    Quantization of Charge

    Total charge Q=neQ = ne
    Q=neQ = ne
  4. 4

    Substitute Values

    Substitute nn and ee into the formula
    Q=(106)×(−1.6×10−19)Q = (10^6) \times (-1.6 \times 10^{-19})
  5. 5

    Calculate Total Charge

    Multiply the values to find QQ
    Q=−1.6×10(6−19)Q = -1.6 \times 10^{(6-19)}
  6. 6

    Final Answer

    Total charge on the rod, Q=−1.6×10−13 CQ = -1.6 \times 10^{-13} \text{ C}
    Q=−1.6×10−13 CQ = -1.6 \times 10^{-13} \text{ C}

Final answer

(b) −1.6×10−13 C-1.6 \times 10^{-13} \text{ C}

Remember: Total charge is integer multiple of elementary charge.

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If a neutral object gains a certain number of electrons, what can be said about its final charge?

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