Question

Particles q1 = -53.0 uc, q2 = +105 uc, and

q3 = -88.0 uc are in a line. Particles qı and q2 are

separated by 0.50 m and particles q2 and q3 are

separated by 0.95 m. What is the net force on

particle qı?

Remember: Negative forces (-F) will point Left

Positive forces (+F) will point Right

-53.0 μC

-88.0 C

+105 με

+92

91

93

K 0.50 m

0.95 m

Enter

no

Answer 1

The answer sir would be 180.38

Step-by-step explanation:

Put in 180.38 trust

Answer 2

`-180.38\ \text{N}`

Step-by-step explanation:

`q_1=-53\ \mu\text{C}`

`q_2=105\ \mu\text{C}`

`q_3=-88\ \mu\text{C}`

r = Distance between the charges

`r_(12)=0.5\ \text{m}`

`r_(23)=0.95\ \text{m}`

`r_(13)=1.45\ \text{m}`

k = Coulomb constant =
`9* 10^9\ \text{Nm}^2/\text{C}^2`

Net force is given by

`F=F_(12)+F_(13)\\\Rightarrow F=(kq_1q_2)/(r_(12)^2)+(kq_1q_3)/(r_(13)^2)\\\Rightarrow F=kq_1((q_2)/(r_(12)^2)+(q_3)/(r_(13)^2))\\\Rightarrow F=9* 10^9* (-53* 10^(-6))((105* 10^(-6))/(0.5^2)+(-88* 10^(-6))/(1.45^2))\\\Rightarrow F=-180.38\ \text{N}`

The force on the particle
`q_1` is
`-180.38\ \text{N}`.