Question

Approximate the field by treating the disk as a +6.1 C point charge at a distance of 21 cm Answer in units of N/C.

Answer 1

Answer:

Electric field, $E=1.24\times 10^{12}\ N/C$

Explanation:

It is given that,

Charge, Q = +6.1 C

Distance, r = 21 cm = 0.21 m

We need to find the electric field. It is given by :

$E=k\dfrac{Q}{r^2}$

$E=9\times 10^9\times \dfrac{6.1}{(0.21)^2}$

$E=1.24\times 10^{12}\ N/C$

So, the electric field at this distance is $1.24\times 10^{12}\ N/C$. Hence, this is the required solution.