Negative electric potential. Phys4 AQA
From my perception of electric potential it is the work done per unit +ve test charge on a +ve test charge when it is moved from infinity to a point P in the field. Is this true only for positive charges giving rise to fields?
Take a negative charge which has a radial field which is directed inwards. Surely no work is done when moving a +ve test charge from infinity to a point inside the field, surely work is done when the charge is moved away from the central charge towards infinity?
So is this why negative potentials are defined to be negative; because the +ve test charge gains potential energy when moving away towards infinity?
Take a negative charge which has a radial field which is directed inwards. Surely no work is done when moving a +ve test charge from infinity to a point inside the field, surely work is done when the charge is moved away from the central charge towards infinity?
So is this why negative potentials are defined to be negative; because the +ve test charge gains potential energy when moving away towards infinity?
Never mind, it just clicked. It increases from the negative side to zero.
Original post by Protoxylic
Never mind, it just clicked. It increases from the negative side to zero.
Yes. Just to confirm.
You can think of negative potential as implying that work has to be done on the (+ive) charge to move it from P to infinity against the attractive force from the negative charge. The work done increases the potential energy of the test charge from a minus value to zero.
Original post by Stonebridge
Yes. Just to confirm.
You can think of negative potential as implying that work has to be done on the (+ive) charge to move it from P to infinity against the attractive force from the negative charge. The work done increases the potential energy of the test charge from a minus value to zero.
You can think of negative potential as implying that work has to be done on the (+ive) charge to move it from P to infinity against the attractive force from the negative charge. The work done increases the potential energy of the test charge from a minus value to zero.
I guess you could also think of it mathematically through calculus as V=kQ/r where k = 1/4piepsilon o and thus dV/dr = -kQ/r^2 and E = kQ/r^2
(edited 9 years ago)
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