A positive charge Q is placed at point O. Is the potential difference (Va-Vb) positive,negative,or zero,if Q is (i)positive,(ii)negative?
Can anyone give the concept behind this?
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1 Answer
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Think of it like this. Electric fields point in the directions that positive charges will tend to go, which is towards a decreasing potential. So, if $Q$ is positive, it will have radial electric field lines that will point along decreasing regions of electrostatic potential; that is, point A will have a higher potential than will point B, and so the potential difference $V_A - V_B$ shall be positive.
The argument for the next situation is similar. I’ll leave that one to you.
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$\begingroup$ i understood,but i couldnot understand Electric fields point in the directions that positive charges will tend to go. $\endgroup$ Nov 18, 2014 at 10:59
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$\begingroup$ can u give the logic when q is negative $\endgroup$ Nov 18, 2014 at 11:00
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1$\begingroup$ In general, that’s just an equivalent statement to saying that they point in the direction of decreasing potential. Positive charges tend to move in directions which will yield a negative potential difference ($\Delta V = V_{\text{final}} - V_{\text{initial}}$), and so I merely included both statements to be instructive. As far as when $q$ is negative, I shall leave that to you. Follow the argument above, but think about what happens when you have a negative charge and what type of electric field it generates. $\endgroup$– kbhNov 18, 2014 at 11:00