- #1

yosimba2000

- 206

- 9

q seems to represent a charge on the capacitor plate and dq seems to represent a separate test charge. If I add a charge to the capacitor plate, I take take the resulting votage and multiply by test charge dq to get the work done on dq. For each additional charge added to the capacitor, I continue to multiply the new voltage by dq. But why should I multiply by dq? The voltage created affects any amount of charge, so why is it restricted to affecting only dq?

For the basic Energy equation E = integral of F(x)dx, this means I am splitting up distance x into pieces of size dx, calculating the force at each segment, multiplying it by the distance dx over which the force acts, and adding everything up. I can see why F(x) is related to dx because each value of F(x) only works across distance dx.