Question

Two protons are maintained at a separation of 973 nm. Calculate the electric potential due to the two particles at the midpoint between them. Then, find the magnitude and direction of the electric field there. potential: V magnitude of field: N/C The direction of field is toward one of the protons. another direction. undetermined.

Answer 1

Answer:

V = 2.95*10^-3 V

E = 0 N/C

Explanation:

In order to calculate the electric potential due to the two protons at the midpoint between them, you use the following:

$V=V_1+V_2=k\frac{q}{r}+k\frac{q}{r}=2\frac{kq}{r}$       (1)

where you have taken into account each contribution to the total electric potential, produced by each proton.

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

q: charge of the proton = 1.6*10^{-19}C

r: distance from the point (at the midway distance between the protons) to one proton = 973nm = 973nm/2 = 486.5nm = 486.5*10^-9m

You replace the values of the parameters in the equation (1):

$V=2\frac{(8.98*10^9Nm^2/C^2)(1.6*10^{-19}C)}{486.5*10^{-9}m}\\\\V=2.95*10^{-3}V$

The electric potential is 2.95*10^-3V

The electric field generated at the midpoint in between the protons is zero, because the electric field generated by each proton has the same magnitude but opposite direction.

E = 0N/C