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The work done in carrying a point charge...

The work done in carrying a point charge form one point to another in an electrostatic field depends on the path along which the point charge is carried.

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Let us consider two points `A` and `B` in an electric field. Let the pootentials at `A` and `B` be `V_(A)` and `V_(B)`, respectively. Now , by the definition of potential difference, the potential difference between two points `B` and `A` is the amount of work done in carrying a unit positive charge from `ArarrPrarrB`,
`(W_(APB))/(q)=V_(B)-V_(A)`
`W_(APB)=q(V_(B)-V_(A))` (i)
For path `ArarrQrarrB`,
`(W_(AQB))/(q)=V_(B)-V_(A)`
or `W_(AQB)=q(V_(B)-V_(A))` (ii)
Since the `RHS` of Eqs. (i) and (ii) is the same.
`W_(APB)=W_(AQB)` (iii)
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Knowledge Check

  • What is the work done in moving a unit positive charge from infinity to that point in electric field called?

    A
    Electric potential
    B
    Potential difference
    C
    Electric current
    D
    Electric circuits

  • The work done in carrying a charge of 5 mu C form a point A to a point B in an electric field is 10mJ . The potential difference (V_(B) - V_(A)) is then

    A
    `+2 kV`
    B
    `-2 kV`
    C
    `+200 V`
    D
    `-200 V`

  • The work done in moving a charge from one point to the other is 20J. If the potential difference between the points is 10V, the charge is:

    A
    4.0C
    B
    2.0C
    C
    0.5C
    D
    1.0C