An insulated neutral conductor gains 5.0 \times 10^{13} electrons during a electrostatic c… — JAMB UTME

An insulated neutral conductor gains $5.0 \times 10^{13}$ electrons during a electrostatic charging process. Given that the magnitude of the elementary charge is $e = 1.6 \times 10^{-19}\text{ C}$ , what is the magnitude of the net charge acquired by the conductor in microcoulombs ( $\mu\text{C}$ )?

Answer: 8 μC

Answer

The magnitude of the net electric charge acquired by the conductor is

8.0\ \mu\text{C}

According to the principle of charge quantization, the total magnitude of charge

Q

acquired by gaining

n

electrons is given by

Q = n e

. Substituting

n = 5.0 \times 10^{13}

and

e = 1.6 \times 10^{-19}\text{ C}

gives

Q = 8.0 \times 10^{-6}\text{ C}

. Expressed in microcoulombs,

8.0 \times 10^{-6}\text{ C} = 8.0\ \mu\text{C}

Step-by-Step Solution

Apply the principle of quantization of electric charge formula

Formula

Q = n e

established

Electric charge is quantized and exists in integer multiples of the elementary charge.

Multiply the number of electrons by the elementary charge value

Q = 8.0 \times 10^{-6}\text{ C}

Calculates total electrostatic charge in base SI units.

Convert the value from Coulombs to microcoulombs

8.0\ \mu\text{C}

The unit

1\ \mu\text{C}

equals

10^{-6}\text{ C}

Key Concept

Quantization of Electric Charge

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