Question regarding electrostatic potentials

[thechairman]

Advertisement - Members don't see this ad

Since electric potential is V = E*d, and since E = k*q/d^2, then

V = k*q/d.

From this equation, it predicts that the electric potential decreases as d increases.

But isn't this wrong from a conceptual point of view? We know that charged particles move from regions of high electric potential to low electric potential, shouldn't the electric potential decrease as the distance between two oppositely charged particles decrease?

 

[artaxerxes]

Since electric potential is V = E*d, and since E = k*q/d^2, then

V = k*q/d.

From this equation, it predicts that the electric potential decreases as d increases.

But isn't this wrong from a conceptual point of view? We know that charged particles move from regions of high electric potential to low electric potential, shouldn't the electric potential decrease as the distance between two oppositely charged particles decrease?

it makes sense, watch your signs

V increases when two positive charges (or negative) are brought together, as work is done to overcome the electrostatic repulsion

V decreases when opposite charges are brought together as work is done by the electrostatic force on the test charge.

Test charge I believe is assumed to be positive, q is the charge of whatever object, and can be + or -

in the case of opposite charges, q= -, so V decreases with Distance.