Question

A 30 nC charge is moved from a point where V = 210 V to a point where V = -90 V . How much work is done by the force that moves the charge?

Answer 1

The pd between the two points is 260V .. (190 - (-70) = 260) 

1V = energy transfer of 1J/C 
Total energy transferred = 30.0^-9C x 260 J/C = 7.80^-6 J .. this is a reduction in EPE as the +charge moves from a higher (+190) to a lower potential (-70), so ∆EPE = - 7.80^-6 J 

WD = energy transferred .. .. ►WD = - 7.80^-6 J 

(Electric field direction is towards the -70V point, so the field force moves the charge)

Answer 2

The work done by the force that moves the charge is -9 x 10⁻⁶ J.

Work done in moving the point charge

The work done in moving the point charge is calculated as follows;

W = qΔV

where;

• q is the charge = 30 nC
• ΔV is the change in potential

W = (30 x 10⁻⁹) x (-90 - 210)

W = ((30 x 10⁻⁹) x (-300)

W = -9 x 10⁻⁶ J

Thus, the work done by the force that moves the charge is -9 x 10⁻⁶ J.

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