Question

How is PE = qV derived using knowledge of F = q1q1/(4pi*e*r^2)

I recently started grade 11, and this concept confuses me, since if we place a charge q on a potential V, i do not completely comprehend what values would need to be substituted which would give me the formula. The closest i have gotten is f = kq1/r^2 thus PE = kq1/r (k = 9*10^9N)

Now i do not understand what values needs to be placed in q1, and what the potential V attempts to tell.

Answer 1

@ganeshie8 @ginblossom @AravindG @Awesome781 @Luigi0210 @Preetha
Please help i have a task i need to complete with this
Thank you.

Answer 2

Potential(V)=-\(\int\limits E dr\)

Potential means the work done in taking a unit charge from infinity to a point in the presence of an Electric field E. If you multiply it by q you get the work done in taking a charge q under the same conditions to that point. This is termed as P.E. of the charge. I hope that clears your doubt.

Answer 3

That actually helps in relating electric potential for PE,
however i want to understand how, using PE = kq1/r and the variables involved in a charge q lying on a potential V can we derive the equation PE = qV
Exactly what would the terms q1 and q2 be in this case, and if the r(distance between particles) is 0, since Q is on the potential V, then how would the equation even work, since
PE = kq1/r would be undefined. How can i derive the equation from the basic expressions given? :/

Answer 4

PE = \(\large \int \limits_{\infty}^{R} \overrightarrow{F} . \overrightarrow{dr} = \int \limits_{R}^{\infty} \overrightarrow{F_{el}} . \overrightarrow{dr} = kq_1q_2 \int \limits_{R}^{\infty} \frac{dr}{r^2} = \frac{kq_1q_2}{R}\)