an electron and a proton are fixed at a separation distance of 897 nm. find the magnitude ???? and the direction of the electric field at their midpoint.



Answer :

The electric field at a midpoint is 1.425 x 10⁴ N/C.

We need to know about the electric fields to solve this problem. The electric field produced by a single-point positive charge is a radial field, whose strength is given by the equation

E = k.Q/r²

where E is the electric field, Q is the charge and r is the radius.

From the question above, we know that

k = 9 x 10⁹ Nm²/C²

Q1 =e = 1.6 x 10¯¹⁹ C

Q2 = e = 1.6 x 10¯¹⁹ C

x = 897 nm = 8.97 x 10¯⁷ m

from midpoint (r = 4.495 x 10¯⁷ m)

The electric field radially outward from proton to electron so that the total electric field between the charges will be added up.

Etotal = E1 + E2

Etotal = k.e/r² + k.e/r²

Etotal = 2. k.e/r²

Etotal = 2 x 9 x 10⁹ x 1.6 x 10¯¹⁹/(4.495 x 10¯⁷)²

Etotal = 1.425 x 10⁴ N/C

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