Question

calculate the electric field due to a charge of -8.0 into 10 raise to minus 8 coulumb at a distance of 5 cm from it. ​

Answer 1

Answer:

Electric field due to a point charge at given distance is

$E = 2.88 \times 10^5 N/C$

Explanation:

Electric field due to a point charge at a distance "r" from it is given as

$E = \frac{kq}{r^2}$

here we know that

q = electric charge

r = distance from the charge

Now plug in the given values in it

$E = \frac{kQ}{r^2}$

$E = \frac{(9\times 10^9)(8 \times 10^{-8})}{0.05^2}$

$E = 2.88 \times 10^5 N/C$

#Learn

Topic : Electric Field

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