A sphere with a radius $\mathbf{1}\mathbf{cm}$ has a charge of $\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{C}$ spreads out uniformly over its surface. What is the magnitude of the electric field due to the sphere at a location $\mathbf{4}\mathbf{cm}$ from the center of the sphere?

The given data is listed below as:

• The value of the charge of the sphere is,$Q=2x{10}^{-9}\mathrm{C}$
• The distance of the electric field from the center of the sphere is,$R=4cm\times \left(\frac{1m}{100cm}\right)=4\times {10}^{-2}m$
• The radius of the sphere is, $r=1cm\times \left(\frac{1m}{100cm}\right)=1\times {10}^{-2}m$

The electric field is referred to as a region that helps an electrically charged particle to exert force on another particle.

The electric field of a sphere is inversely proportional to the radius of an object and directly proportional to the charge of that object.

The equation of electric field inside a sphere of uniform charge is expressed as:

$E=k\frac{Q}{R^2}$

Here, $Q$is the total charge, $R$is the distance of the electric field from the center of the sphere and $k$is the electric field constant with value $9\times {10}^{9}N.m^2/C^2$.

Substitute all the values in the above equation.

$$\begin{gathered} E=\left(9\times {10}^{9}N.m^2/C^2\right)\frac{2\times {10}^{-9}C}{{\left(4\times {10}^{-2}m\right)}^2} \\ =\frac{18N.m^2/C}{1.6\times {10}^{-3}{m}^2} \\ =1.12\times {10}^{4}N/C \end{gathered}$$

Thus, the magnitude of the electric field is $1.12\times {10}^{4}N/C$.