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?

Question

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

are we ignoring the test charge (q)?

unklerhaukus · Answer

maybe the test charge is one unit