Question

Approximate the field by treating the disk as a +6.7 μ C point charge at a distance of 48 cm.

Answer 1

Answer:

$E = 2.62 \times 10^5 N/C$

Explanation:

Electric field due to a point charge is given as

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

here we know that

$q = 6.7 \mu C$

$r = 48 cm$

now plug in all values in it

$E = \frac{(9\times 10^9)(6.7\mu C)}{0.48^2}$

$E = 2.62 \times 10^5 N/C$

so here electric field is given due to point charge as above