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Difficulty: EasyElectrostatics and Electric Charges

A neutral insulated conductor loses 5.0×10135.0 \times 10^{13} electrons during an electrostatics experiment. What is the magnitude of the net electric charge, in microcoulombs (μC\mu\text{C}), acquired by the conductor? (Take elementary charge e=1.6×10−19 Ce = 1.6 \times 10^{-19}\text{ C})

Answer: 8 μC

Answer

The magnitude of the net electric charge acquired by the conductor is 8 μC8\text{ }\mu\text{C}.
According to the principle of charge quantization, the total electric charge QQ is calculated using Q=neQ = n e. Multiplying 5.0×10135.0 \times 10^{13} electrons by 1.6×10−19 C1.6 \times 10^{-19}\text{ C} yields 8.0×10−6 C8.0 \times 10^{-6}\text{ C}, which converts to 8 μC8\text{ }\mu\text{C}.

Step-by-Step Solution

1
Identify the relevant formula for quantization of charge.
The net charge acquired is given by Q=neQ = n e.
Electric charge is quantized, so the total charge magnitude equals the number of transferred electrons multiplied by the magnitude of charge on a single electron.
2
Calculate the magnitude of charge in Coulombs.
Q=(5.0×1013)×(1.6×10−19 C)=8.0×10−6 CQ = (5.0 \times 10^{13}) \times (1.6 \times 10^{-19}\text{ C}) = 8.0 \times 10^{-6}\text{ C}.
Multiplying the quantity of removed electrons by the elementary charge gives total charge in Coulombs.
3
Convert the calculated value from Coulombs to microcoulombs.
8.0×10−6 C=8 μC8.0 \times 10^{-6}\text{ C} = 8\text{ }\mu\text{C}.
Since 1 μC=10−6 C1\text{ }\mu\text{C} = 10^{-6}\text{ C}, dividing 8.0×10−68.0 \times 10^{-6} by 10−610^{-6} yields 8.

Key Concept

Quantization of Electric Charge
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Topics

Electrostatics and Electric Charges

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