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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?
 one year ago
 one year 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?
 one year ago
 one year ago

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JenniferSmart1Best ResponseYou've already chosen the best response.0
\[\Delta V=V_bV_a=\frac{\Delta U}{q_0}=\int_a^b\vec E \cdot d\vec l\]
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
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
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
are we ignoring the test charge (q)?
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
maybe the test charge is one unit
 one year ago
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