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Worked Examples · Example 1.1

Q.If 10910^9 electrons move out of a body to another body every second, how much time is required to get a total charge of 1 C1\,\text{C} on the other body?

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✓ Free question

Charge accumulates at a rate determined by the number of electrons transferred per second. At 10910^9 electrons/second, it takes approximately 200 years to accumulate 1 C1\,\text{C}.

Understanding charge transfer

When electrons move from one body to another, they carry their elementary charge e=1.6×10−19 Ce = 1.6 \times 10^{-19}\,\text{C} with them. The body losing electrons becomes positively charged, while the body gaining electrons becomes negatively charged. The question asks how long it takes for the receiving body to accumulate a total charge of 1 C1\,\text{C}.

The key insight is that charge builds up at a constant rate if electrons transfer at a steady rate. We need to find the charge transferred per second (the current), then determine how many seconds are needed to reach 1 C1\,\text{C}.

Step-by-step solution

1. Calculate the charge transferred per second

Each electron carries charge e=1.6×10−19 Ce = 1.6 \times 10^{-19}\,\text{C}. If 10910^9 electrons move every second, the charge transferred per second is:

I=n⋅e=109×1.6×10−19 C/sI = n \cdot e = 10^9 \times 1.6 \times 10^{-19}\,\text{C/s}

I=1.6×10−10 C/sI = 1.6 \times 10^{-10}\,\text{C/s}

This is effectively the current flowing between the two bodies.

2. Find the time required to accumulate 1 C1\,\text{C}

We know that charge Q=I⋅tQ = I \cdot t, where tt is time. Rearranging for time:

t=QI=1 C1.6×10−10 C/st = \frac{Q}{I} = \frac{1\,\text{C}}{1.6 \times 10^{-10}\,\text{C/s}}

t=11.6×10−10 s=10101.6 st = \frac{1}{1.6 \times 10^{-10}}\,\text{s} = \frac{10^{10}}{1.6}\,\text{s}

t=6.25×109 st = 6.25 \times 10^9\,\text{s}

3. Convert to more meaningful units

This is an enormous amount of time. Let's convert to years:

t=6.25×109365×24×3600 yearst = \frac{6.25 \times 10^9}{365 \times 24 \times 3600}\,\text{years}

t=6.25×1093.156×107 yearst = \frac{6.25 \times 10^9}{3.156 \times 10^7}\,\text{years}

t≈198 yearst \approx 198\,\text{years}

Watch out

A coulomb is a huge amount of charge at the scale of individual electrons. Even though 10910^9 sounds like a large number, the elementary charge is so tiny that the transfer rate is extremely slow. This is why we rarely see static charges of 1 C1\,\text{C} in everyday life!

Tip

Remember that 1 A=1 C/s1\,\text{A} = 1\,\text{C/s}. The current here is only 1.6×10−10 A1.6 \times 10^{-10}\,\text{A}, which is a billionth of an ampere — far smaller than typical household currents of a few amperes.

✓Final answer

The time required is 6.25×109 s\boxed{6.25 \times 10^9\,\text{s}} or approximately 198 years.

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