Find the electric potential vp at point p. [hint: to input a natural logarithm into the answer box, simply type the letters "ln" (for example ln(x) is the natural logarithm of x).] express your answer in terms of d, l, q, and ϵ0.

Iwannasay c is the most accurate

stands for electromotive force and it is measured in volts. despite its name, it is not actually the force in the sense of being a "pushing" or "pulling" (not measured in newtons).an equation that relates emf,  ℰ,  to a battery's terminal voltage, v, current, i, and internal resistance, r, is    ℰ = v + irin this case, the emf can be interpreted as a voltage—a difference in electric potential.i would not say that it would be considered an energy. if emf can be defined as the energy/work per a unit charge from a battery or other electric source, then it is not strictly energy.

0 = 0.40 / 0 = 0

= 0.20 = 9.80 ⁄2 =?

=0 , = = 0.20 =0.50 0 0.40 ⁄

=0−122 =0−12(9.80 ⁄2)(0.50)2 =1.2