Question

The work done in moving 4 μC charge from one point to another in an electric field is 0.012 J. The potential difference between them is

Answer 1

Given :

➠ Magnitude if charge = 4uC

➠ Work done = 0.012J

To Find :

➛ Pd between those points.

SoluTion :

✭ Work done in moving a charge from one point to another point in an uniform electric field is given by

• W = q × ΔV

where,

➝ W denotes work done

➝ q denotes charge

➝ ΔV denotes pd

⇒ W = q × ΔV

⇒ 0.012 = (4 × 10$^{-6}$) × ΔV

⇒ ΔV = 0.012/4×10$^{-6}$

⇒ ΔV = 3000 volt

Answer 2

Answer:

  3000 V    OR      3.0 * 10³  V

Explanation:

  Charge, Q = 4μC   =  (4*10⁻⁶)C

  Work done, W = 0.012J

  Potential difference, V = ?   (to be calculated)

We know that,

             V = W / Q

 Putting the above values,

             V = 0.012 / 4*10⁻⁶

                 = 0.003 * 10⁶

                 = 3000 V

                        OR

                 = 3.0 * 10³ V

Hope you are happy with my solution... Study hard!!!