Question

Two point charges, each of magnitude 20 uC are separated by a distance of 1.5 m in a large vessel of
water. The relative permittivity of water is 80. Calculate the force between them​

Answer 1

Answer:

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Answer 2

The force between charges is $1.99\times 10^{-2}\ N$

Explanation:

Two point charges, each of magnitude $20\ \mu C$ are separated by a distance of 1.5 m in a large vessel of  water.

The relative permittivity of water is 80. The relative permittivity is given by :

$k=\dfrac{\epsilon}{\epsilon_o}$

New force becomes :

$F=\dfrac{1}{4\pi \epsilon_o k}\dfrac{q_1q_2}{r^2}\\\\F=\dfrac{1}{4\pi\times 8.85\times 10^{-12}\times 80}\dfrac{(20\times 10^{-6})^2}{(1.5)^2}\\\\F=0.0199\ N\\\\F=1.99\times 10^{-2}\ N$

So, the force between charges is $1.99\times 10^{-2}\ N$. Hence, this is the required solution.

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