Question

Two charges are in the configuration indicated here. The first charge, Q1 = –1.00 μC, sits at the origin. The second charge, Q2 = +1.00 μC, sits at (x, y) = (1.00 m, 1.00 m). What is the the electric field strength at point P, the midpoint between the two charges?

Answer 1

Answer:

$E_{net} = 3.6 \times 10^4 N/C$

Explanation:

As the two charges Q1 and Q2 are placed at some distance apart

so the electric field at mid point will be twice the electric field due to one charge

Because here the two charges are of opposite sign so here the electric field at mid point will be added due to both

so here we have

$E_{net} = 2E$

$E_{net} = 2(\frac{kQ}{r^2})$

distance of mid point from one charge is given as

$r = \frac{\sqrt{1^2 + 1^2}}{2}$

$E_{net} = 2 (\frac{(9\times 10^9)(1\times 10^{-6})}{(\frac{1}{\sqrt2})^2}$

$E_{net} = 3.6 \times 10^4 N/C$