Question

A 100 nC charge is placed at the origin. A -50 nC charge is placed at x= 5 cm and a 200 nC is placed at x= -10 cm.

a. What is the net electrostatic force acting on 100 nC charge?

Answer 1

Solution.

$q_1=100\sdot10^{-9}C;$

$q_2=-50\sdot10^{-9}C;$

$q_3=200\sdot10^{-9}C;$

$x_2=5cm=0.05m;$

$x_3=-10cm=-0.1m;$

$F-?;$

$F_{21}=k\dfrac{q_2q_1}{r_2^2};$

$F_{21}=9\sdot10^9\dfrac{50\sdot10^{-9}\sdot100\sdot10^{-9}}{0.05^2}=1.8\sdot10^{-2}N;$

$F_{31}=k\dfrac{q_3q_1}{r_3^2};$

$F_{31}=9\sdot10^9\dfrac{200\sdot10^{-9}\sdot100\sdot10^{-9}}{0.1^2}=1.8\sdot10^{-2}N;$

$F=F_{21}-F_{31};$

$F=1.8\sdot10^{-2}-1.8\sdot10^{-2}=0N;$

Answer: $F=0N.$