Electrostatics question (1 Viewer)

erucibon · Aug 29, 2019

Calculate electric field strength due to the 4μ C charge at the position of the other charge

Thanks

Pedro123 · Aug 29, 2019

By Coloumb's law: (Apologies I cannot write equations)
The force felt would be (Coloumb's Constant*absolute value(Charge 1 * Charge 2))/distance^2
Coloumb's constant is about 8.987*10^9 (Units of NM^2C^-2
Subbing in, we get a value of about 7.99x10^11 newtons

jazz519 · Aug 29, 2019

Pedro123 said:
By Coloumb's law: (Apologies I cannot write equations)
The force felt would be (Coloumb's Constant*absolute value(Charge 1 * Charge 2))/distance^2
Coloumb's constant is about 8.987*10^9 (Units of NM^2C^-2
Subbing in, we get a value of about 7.99x10^11 newtons

It's asking for the electric field strength rather than the force between the particles

So you need to use that formula but without the other charge (so E = kq1/r^2)

Drdusk · Aug 29, 2019

Yeah

$F = q_2E = \frac{1}{4\pi \epsilon_0}\frac{q_1q_2}{r^2}$

$\therefore E = \frac{1}{4\pi \epsilon_0}\frac{q_1}{r^2}$

$\text{Subbing in we get:}$

$E = \frac{1}{4\pi \epsilon_0}\frac{4\times 10^{-6}}{(0.6)^2} = 99864 NC^{-1}$

$\approx 100000 NC^{-1}$

Electrostatics question (1 Viewer)

erucibon

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