Question

Two charges are separated by 1.68 cm. Object A has a charge of 5.0 μ C , while object B has a charge of 7.0 μ C . What is the force on Object A?

Answer 1

Answer:

The force on Object A = 1116.07 N

Step-by-step explanation:

The separation between the two charges, d = 1.68 cm

d = 1.68/100 m

d = 0.0168 m

The charge on object A

`\begin{gathered} q_A=5.0\mu C \\ q_A=5*10^(-6)C \end{gathered}`

The charge on object B

`\begin{gathered} q_B=7.0\mu C \\ q_B=7*10^(-6)C \end{gathered}`

The electric constant

`k=9*10^9Nm^2C^(-2)`

The force on on the charge A is calculated as shown below:

`\begin{gathered} F_A=(kq_Aq_B)/(d^2) \\ F_A=(9*10^9*5*10^(-6)*7*10^(-6))/(0.0168^2) \\ F_A=(0.315)/(0.00028224) \\ F_A=1116.07N \end{gathered}`

The force on Object A = 1116.07 N