Author Topic: Electrostatics question (Read 1034 times) Tweet Share

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

fun_jirachi · « **Reply #1 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

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Author Topic: Electrostatics question (Read 1034 times) Tweet Share

erucibon

Electrostatics question

fun_jirachi

Re: Electrostatics question

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