Question

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.)

Short answer

The maximum amount of charge you can place on the sphere is$\begin{array}{rcl}& & 7.5\times {10}^{-6} \text{C}\end{array}$

Step-by-step solution

Identification of given data

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

Significance of electric field

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

Determining the maximum amount of charge 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" $\begin{array}{rcl}& & 7.5\times {10}^{-6} \text{C}\end{array}$