## anonymous 3 years ago How does $$\Delta V$$ and $$\Delta U$$have the same magnitude for a given charge, when the change in U the integral multiplied by Q?

1. anonymous

$\Delta V=V_b-V_a=\frac{\Delta U}{q_0}=-\int_a^b\vec E \cdot d\vec l$

2. anonymous

we'll I'm being told that "changes in U and V have the same magnitude for a given charge; they only depend on E.dl (the dot product of each differential path vector with the electric field)" and I don't understand how U and V can have the same magnitude

3. anonymous

are we ignoring the test charge (q)?

4. UnkleRhaukus

maybe the test charge is one unit