An automobile of mass 1.46 cross times to the power of blank 10 cubed kg rounds a curve of radius 25.0 m with a velocity of 15.0
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Answer:
The electric potential is approximately 5.8 V
The resulting direction of the electric field will lie on the line that joins the charges but since it is calculated in the midpoint and the charges are the same we can directly say that its magnitude is zero
Explanation:
The two protons can be considered as point charges. Therefore, the electric potential is given by the point charge potential:
(1)
where is the charge of the particle,
the electric permittivity of the vacuum (I assuming the two protons are in a vacuum) and
is the distance from the point charge to the point where the potential is being measured. Because the electric potential is an scalar, we can simply add the contribution of the two potentials in the midpoint between the protons. Thus:
Substituting the values ,
and
we obtain:
The resulting direction of the electric field will lie on the line that joins the charges but since it is calculated in the midpoint and the charges are the same we can directly say that its magnitude is zero.