Answer to Question #120439 in Physics for nihar

Question

A +1.6nC point charge is placed at one corner of a square (2.00 m on a side), and a -2.40nC charge is placed on the corner diagonally opposite. What is the magnitude of the electric field intensity at either of the other two corners? What would be an acceleration of a particle with a charge of +1nC placed at the other corner of this square?

Answer 1

1)

$$E_x=k\frac{q_1}{a^2}=8.99\cdot10^{9}\frac{1.6\cdot10^{-9}}{2^2}=3.596\frac{N}{C}$$

$$E_y=k\frac{q_2}{a^2}=8.99\cdot10^{9}\frac{-2.4\cdot10^{-9}}{2^2}=-5.394\frac{N}{C}$$

$$E=\sqrt{3.596^2+(-5.394)^2}=6.48\frac{N}{C}$$

2)

$$a=\frac{F}{m}$$

$$F=qE=(1\cdot10^{-9})(6.48)=6.48\cdot10^{-9}N$$