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JenniferSmart1
 2 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?
JenniferSmart1
 2 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?

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JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0\[\Delta V=V_bV_a=\frac{\Delta U}{q_0}=\int_a^b\vec E \cdot d\vec l\]

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0we'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

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0are we ignoring the test charge (q)?

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0maybe the test charge is one unit
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