Question

A particle of charge 5×10^-2 coulomb is taken from a point at a potential of 50v to another point at a potential of 250v.calculate the work done

Answer 1

Solution :

⏭ Given:

✏ A particle of charge 0.05C is taken from a point at a potential of 50V to another point at a potential of 250V.

⏭ To Find:

✏ Work done

⏭ Formula:

✏ Formula of work done in terms of charge and potential difference is given by

• W = q × ΔV

⏭ Terms indication:

• W denotes work done
• q denotes charge on particle
• ΔV denotes p.d.

⏭ Calculation:

• W = 0.05 × (250-50)
• W = 0.05 × 200
• W = 10J

✏ Net work done to move a charge particle = 10J

Answer 2

$\mathtt{ \huge{ \fbox{Solution :)}}}$

Given ,

• Potential difference = (250 - 50) volt
• Charge (q) = 6 C

We know that , the work done (w) to move a unit charge (q) from infinity to a particular point is called potential difference

$\large \mathtt{\fbox{Potential \: difference = \frac{w}{q} } }$

Thus ,

200 = Work done/5 × (10)^-2

Work done = 1000 × (10)^-2

Work done = 10 joule

Hence , the work done is 10 volt

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