That sounds right. You need to find the change in electric potential energy as the charges are brought together.rojasharma said:what i did was..used Ee1=kq1q2/r1 (r1 as 0.4m), Ee2=kq1q2/r2(r2 as 100m), then found delta Ee= Ee2=Ee1, Delta Ee= W??
Looks OK. (You seemed to have switched r1 & r2: the charge moves from 100m to 0.4m.)rojasharma said:So,
q1=3.2x10^-3C
q2=1x10^-6C
r1=0.4m
r2=100m
Solution
E1= 72J (E1= (kq1q2)/r1)
E2= 0.288J (E2=kq1q2/r2)
Where did you get 180J from? The negative sign is due to you mixing up initial and final positions.delta E= E2-E1
delta E= 0.288J - 180J
delta E=-179.712J
Delta E= work done...(what about the negative value??)
Doc Al said:That sounds right. You need to find the change in electric potential energy as the charges are brought together.
Which is how one would derive the expression for electric potential energy.PaintballerCA said:I am thinking of writing the force between the two charges and integrating it over the distance.
The work done to move a charge is the amount of energy required to move a charged particle from one point to another point in an electric field.
The unit of measurement for work done to move a charge is joules (J).
Work done to move a charge is calculated by multiplying the magnitude of the charge (in coulombs, C) by the potential difference (in volts, V) through which it moves. The equation for work is W = QV.
Yes, the direction of the charge's movement does affect the work done. Work is a scalar quantity, so it is only affected by the magnitude of the charge and potential difference. However, the direction of the charge's movement can affect the sign of the work, indicating whether work is being done on or by the charge.
The relationship between work done and potential difference is directly proportional. This means that as the potential difference increases, the work done to move a charge also increases. Similarly, as the potential difference decreases, the work done to move a charge decreases.