Question

Under the influence of electrostatic field of an unknown system of fixed charges, a 2C charge is moved from point C to D slowly. The change in potential energy of the system is 10 J. What is the work done by the mentioned conservative field in the process?

• A. -10J
• B. 10J
• C. Zero
• D. 20J

Answer 1

The correct option is A -10J

The change in potential energy of a system is the work done by an external agent in moving the charge slowly (slowly means -without acceleration) from one point to another (in this case C to D)
Work done by external agent = Change in potential energy = 10J (given) ………………………..(1)
We also know that electrostatic field (i.e. field created by static charges) is conservative and from the work energy theorem we have
Change in kinetic energy = Net work done on the system
Here, change in kinetic energy is zero as there was no acceleration (hence no change in velocity)

So the net work done in moving a charge from C to D must also be zero according the work energy theorem. If that is the case and let’s say the work done by external agent is W_extthen the work done by field W _field must be equal and opposite to it
$W_{field}=-W_{ext}$
From (1) we have,
$W_{ext}=10J$
So $W{­}_{­field}=-10J$