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Mathematics 13 Online
OpenStudy (anonymous):

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?

OpenStudy (anonymous):

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

OpenStudy (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

OpenStudy (anonymous):

are we ignoring the test charge (q)?

OpenStudy (unklerhaukus):

maybe the test charge is one unit

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