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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?

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  1. JenniferSmart1
    • 2 years ago
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    \[\Delta V=V_b-V_a=\frac{\Delta U}{q_0}=-\int_a^b\vec E \cdot d\vec l\]

  2. JenniferSmart1
    • 2 years ago
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    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. JenniferSmart1
    • 2 years ago
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    are we ignoring the test charge (q)?

  4. UnkleRhaukus
    • 2 years ago
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    maybe the test charge is one unit

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