Question

Calculate the electric field intensity at a point at 2 m distance from a 5 C charge in water.
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Answer 1

Answer:

Explanation:

E=kq/r2

k= 9x10^9

E=( 9X10^9X5)/2

E=202.5X10^9

Answer 2

The answer is 1.46*10^8.

Given:

A 5 C charge in water

To Find:

The electric field intensity at a 2 m distance from the charge.

Solution:

The electric field intensity is given by

$\frac{q}{4\pi \epsilon r^2}$

Here q = 5 C, r = 2 m

∈ for water is $80*8.85*10^{-12}=6.8*10^{-10}$

Therefore the electric field intensity is

$\frac{5}{4\pi *6.8*10^{-10}*4} =1.46*10^8$

The electric field intensity is $\bf1.46*10^8$.

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