Question

What will be the magnitude of the electrical field created by 4.95 * 10 ^ - 4 charge particle 7.01 * 10 ^ - 2 away?

Answer 1

The magnitude E of the electrical field created by a charge particle with charge q at a distance r away, is:

`E=(kq)/(r^2)`

Where k is the Coulomb constant which has a value:

`k=8.99*10^9N\cdot(m^2)/(C^2)`

Substitute q=4.95*10^-4 C and r=7.01*10^-2 m to find the magnitude of the electrical field:

`\begin{gathered} E=(kq)/(r^2) \\ =\frac{(8.99*10^9N\cdot(m^2)/(C^2))(4.95*10^(-4)C)^{}}{(7.01*10^(-2)m)^2} \\ =(4.45*10^6(N)/(C)m^2)/(4.91*10^(-3)m^2) \\ =9.06*10^8\cdot(N)/(C) \end{gathered}`

Therefore, the answer is option b:

`9.08*10^6(N)/(C)`