Question

Two charges, +4 µC and +14 µC, are fixed 1 m apart, with the second one to the right. Find the magnitude and direction of the net force (in N) on a −5 nC charge when placed at the following locations.

Answer 1

`0.0018\ \text{N/C}` towards the right.

`0.001\ \text{N/C}` towards the right.

Step-by-step explanation:

`q_1=4\ \mu\text{C}`

`q_2=14\ \mu\text{C}`

`Q=5\ \text{nC}`

`r_1=r_2=0.5\ \text{m}`

Let
`Q` be placed at origin so
`q_1` becomes negative and
`q_2` becomes positive

Electric field is given by

`E=(kq_1Q)/(r_1^2)+(kq_2Q)/(r_2^2)\\\Rightarrow E=(kQ)/(r^2)(q_1+q_2)\\\Rightarrow E=(9*10^(9)* 5*10^(-9))/(0.5^(2))(-4*10^(-6)+14*10^(-6))\\\Rightarrow E=0.0018\ \text{N/C}`

The electric field halfway between the points is
`0.0018\ \text{N/C}` towards the right.

`r_1=0.5\ \text{m}`

`r_2=1+0.5=1.5\ \text{m}`

`E=9* 10^9* 5* 10^(-9)((4* 10^(-6))/(0.5^2)+(14* 10^(-6))/(1.5^2))\\\Rightarrow E=0.001\ \text{N/C}`

The electric field halfway between the points is
`0.001\ \text{N/C}` towards the right.