Question

Question 3 of 5What could you do to increase the electric force between two chargedparticles by a factor of 16?A. Reduce one particle's charge by a factor of 16.B. Reduce one particle's charge by a factor of 4.C. Increase one particle's charge by a factor of 16.O D. Increase one particle's charge by a factor of 4.

Answer 1

Step-by-step explanation

the force between 2 charges is given by the formula:

`\begin{gathered} F=(Kq_1q_2)/(r^2) \\ where \\ q_i\text{ is the charge on object i} \\ r\text{ is the distance between the charges} \\ K=Coulomb\text{ constant=8.9875*10}^9N*(m^2)/(c^2) \end{gathered}`

then,we can check every option and compare the result to the original, hence

Step 1

a)option A

reduce one particle's charge by a factor of 16

so

`q^(\prime)_1=(q_1)/(16)`

now, replace

`\begin{gathered} F=(Kq_1q_2)/(r^2) \\ F_A=(K(q_1)/(16)q_2)/(r^2)=(1)/(16)(Kq_1q_2)/(r^2) \\ F_A=(1)/(16)(original) \\ F_A=(1)/(16)(Kq_1q_2)/(r^2) \end{gathered}`

therefore, option A works

Step 2