is zero. That is, work to be done against electrostatic force to move a unit test charge from
B to A over a path outside the source + work to be done against electrostatic force to move
a unit test charge from A to B over a path inside the source is zero. Therefore, work to be
done in path B–O–Awork to be done in path B–I–A. But the electrostatic field vector
inside the source is equal to –. Therefore,
But is the work done bythe non-electrostatic force generated by the
source on a unit positive charge when it moves through the source from negative terminal
to positive terminal. This quantity is defined as the Electromotive force (e.m.f.) of the ...
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