The correct expression for the potential at a point distance r from a charge q in an electric field is?
| _q2 |
| 4πεor2 |
| _q2 |
| 4πεor |
| _q |
| 4πεor2 |
\(\frac{q}{4πε_or}\)
Explanation
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Discussions (15)
You guys are making a mistake
Electric Field Intensity=q/4π£or^2
But the question asked for electric potential which is q/4π£or
Therefore they are actually correct
The options are incorrect. Option d is closest to the answer but the radius doesn't possess a square of which the original expression does.
Electric potential at any point in an electric field is the work done in bringing a unit positive charge from infinity to that point against electrical forces of the field
V=Work done/Charge
but work done=force * distance FD
therefore V=FD/q
but force in an electric field= q1q2/4π£r²
and distance= r
therefore V = q1q2/4π£r² * r / q
q1=q2=q
therefore V=q²*r / 4π£r² * q
r² divide r = r
q² divide q = q
V= q / 4π£r
I think for potential difference we use diameter. and diameter is 2r. so we input it instead of just using r
Option d isn't the correct answer, the radius is meant to be squared. There is no correct option. Option c would have been correct but the q isn't negative. Thank you.
