Explain briefly why there is a limit to how much charge can be placed on a metal sphere in the classroom. If the radius of the sphere is $\mathbf{15}\mathbf{ }\mathbf{\text{cm}}$ what is the maximum amount of charge you can place on the sphere? (Remember that a uniform sphere of charge makes an electric field outside the sphere as though all the charge were concentrated at the center of the sphere.)

Radius of sphere is $r=15 \text{cm} \text{or} \text{0.15} \text{m}$

The electric force per unit charge is referred to as the electric field. It is assumed that the field's direction corresponds to the force it would apply to a positive test charge. From a positive point charge, the electric field radiates outward, and from a negative point charge, it radiates in.

$E=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2}$ ...(i)

Where, ${\epsilon }_0$is permittivity of the free space ( $8.85\times {10}^{-12} {\text{C}}^{\text{2}}{\text{/Nm}}^{\text{2}}$)

$q$is the maximum amount of charge that you can place on the sphere

Using equation (i)

$\begin{array}{rcl}E& =& \frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2}\\ q& =& E\left(4\pi {\epsilon }_0{r}^2\right)\end{array}$

Electric field required to cause a spark in air at STP is$E=3\times {10}^6 \text{V/m}$

Substitute all the values in above equation,

$\begin{array}{rcl}q& =& \left(3\times {10}^6 \text{V/m}\right)\times 4\times \pi \times \left(8.85\times {10}^{-12} {\text{C}}^{\text{2}}\text{/Nm}\right)\times {\left(\text{0.15} \text{m}\right)}^2\\ & =& 7.5\times {10}^{-6} \text{C}\end{array}$

Hence the maximum amount of charge you can place on the sphere islocalid="1662209970981" $\begin{array}{rcl}& & 7.5\times {10}^{-6} \text{C}\end{array}$