Question

What are the strength and direction of the electric field 4.0 cm from a small plastic bead that has been charged to -8.0 nC?

Short answer

The strength of the electric field is$-4.5\times {10}^{4}N/C$

Step-by-step solution

Given information

electron distance $d=4.0\times {10}^{-2}m$

the charge of an electron is given by$=-8\times {10}^{-9}C$

Explanation 

The electric fieldis defined as the electric force per unit charge.

$\overrightarrow{E}=\frac{\overrightarrow{F^{\mathrm{\prime }}}}{q}$

We want to find the electric field at a distance $d=4.0\times {10}^{-2}m$. The electric field of a single point charge is given by $E=k_e\frac{Q}{d^2}$

$\text{where}{k}_e=8.99\times {10}^9{\mathrm{Nm}}^2{\mathrm{C}}^{-2}$

So, substituting, we find the electric field at the distance d:

$E=\left(8.99\times {10}^9{\mathrm{Nm}}^2{C}^{-2}\right)\left(\frac{-8.0\cdot {10}^{-9}C}{(0.04m{)}^2}\right)=-4.5\times {10}^4\mathrm{N}/\mathrm{C}$

Therefore, the strength of the electric field is$-4.5\times {10}^{4}N/C$