Question

What is the magnitude of the electric field 17.1 cm directly above an isolated 1.83Ã10â5 C charge?

Answer 1

Answer:

Electric field, $E=5.63\times 10^{16}\ N/C$

Explanation:

Given that,

Charge, $q=1.83\times 10^5\ C$

We need to find the magnitude of electric field 17.1 cm (0.171 m) above an isolated charge. Electric field at a point is given by :

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

$E=\dfrac{9\times 10^9\times 1.83\times 10^5\ C}{(0.171\ m)^2}$

$E=5.63\times 10^{16}\ N/C$

So, the electric field is $5.63\times 10^{16}\ N/C$. Hence, this is the required solution.