ATAR Notes: Forum

HSC Stuff => HSC Science Stuff => HSC Subjects + Help => HSC Physics => Topic started by: erucibon on August 29, 2019, 09:25:55 am

Title: Electrostatics question
Post by: erucibon on August 29, 2019, 09:25:55 am: Calculate electric field strength due to the 4μ C charge at the position of the other charge

Thanks
Title: Re: Electrostatics question
Post by: fun_jirachi on August 29, 2019, 10:09:37 am: Hey there!

The force between the two charges is given by the formula below:
$F=\frac{1}{4\pi \epsilon _0}\frac{q_1q_2}{r^2}$
We also know that F=Eq, so therefore, at the position of the 8µC charge, the field strength due to the 4µC charge can be calculated by
$E=\frac{\frac{1}{4\pi \epsilon _0}\frac{q_1q_2}{r^2}}{q} = \frac{\frac{1}{4\pi \epsilon _0}\frac{4 \times 10^{-6} \times 8 \times 10^{-6}}{0.6^2}}{8 \times 10^{-6}} = 10000 \ \text{N C}^{-1}$

Hope this helps :)