Question

The electric potential at the 2 ends of conductor are 10V and 5V respectively.
5 J of work needs to be to move certain amount of charge from the end at the higher potential to lower potential. Find the amount of charge moved..??

Answer 1

Answer:

potential difference is = 10 - 5 = 5V

V = W/Q ( Q is charge, W is potential difference)

so Q = W/V

Q= 5/5

Q = 1

amount of charge moved is 1 coulumb.

Answer 2

Therefore the amount of charge moved from the end at the higher potential to the lower potential is equal to -1 Coulomb.

Given:

Electric potential of the 1st end of the conductor (Higher Potential) = 10 V

Electric potential of the 2nd end of the conductor (Lower Potential) = 5 V

Work done in moving the charge = 5 J

To Find:

The amount of charge moved (Q)

Solution:

→ Potential difference between the 1st and 2nd end (ΔV) = 5 - 10 = - 5V

                                            ∴ ΔV = -5 V

→ The potential difference (ΔV) between two points is equal to the amount of work done by an external agent in moving a unit positive charge from one point to the other.

→ Since the amount of work done by external agent in moving unit positive charge from one point to the other is ΔV, therefore the amount of work done by an external agent in moving the 'Q' amount of charge from one point to the other would be equal to Q(ΔV).

                                ∴ Work Done (W) = Q(ΔV)

                                     ∵ W = Q × ΔV = Q × 5

                                     ∴ 5 = -5 × Q

                                     ∴ Q = -1 Coulomb

Therefore the amount of charge moved from the end at the higher potential to lower potential is equal to -1 Coulomb.

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