Question

What is the strength of the electric field 3.0 cm from a small glass bead that has been charged to + 8.0 nc ?

Answer 1

Thinking the small glass bead as a single point charge, the electric field generated by it is given by

`E(r) = k_e (Q)/(r^2)`
where

`k_e = 8.99 \cdot 10^9 N m^2 C^(-2)` is the Coulomb's constant

`Q=8.0 nC=8.0 \cdot 10^(-9) C` is the charge of the bead

`r=3.0 cm=0.03 m` is the distance at which we calculate the field.

Using these data, we find:

`E=(8.99 \cdot 10^9 N m^2 C^(-2) ) ((8.0 \cdot 10^(-9) C)/(0.03 m)^2)=8.0 \cdot 10^4 N/C`