Question

A charge of 6micro coulomb is brought from a point x to point y such that the potential difference between the points is 25 volts. What is the work done in moving the charge?

Answer 1

As we know,

Potential difference = workdone/ charge

$25 = \frac{work \: done}{6 \times {10}^{ - 6} } \\ \\ work \: done = 25 \times 6 \times {10}^{ - 6} \\ = 150 \times {10}^{ - 6}joules \\ =1.5 \times {10}^{ - 4} joules$

Answer 2

Answer:

Explanation:

We know that potential difference = Work done/Charge moved.

The potential difference = 25V

The amount of charge moved = 6μC = 6 × 10^-6 C

Therefore, Work done = Potential difference × Charge moved.

I.e. 25 × 6 × 10^-6 J

= 150 × 10^-6 J

= 150 μJ