A proton at speed v = 3×10⁵m/sorbits at radius r = 1.00 cmoutside a charged sphere. Find the sphere’s charge.

Speed, $\mathrm{v}=3.0\times {10}^5\mathrm{m}/\mathrm{s}.$

Radius, r = 1.00 cm.

The proton is in a uniform circular motion, with the electrical force of the sphere on the proton providing the centripetal force. According to Newton’s second law:

$\mathbf{F}\mathbf{}\mathbf{=}\mathbf{}\frac{{\mathbf{mv}}^{\mathbf{2}}}{\mathbf{r}}$

Where F is the magnitude of the force, m is the mass, v is the speed of the proton, and r is the radius of its orbit, essentially the same as the radius of the sphere.

The magnitude of the force on the proton is:

$\mathbf{F}\mathbf{=}\frac{\mathbf{e}\mathbf{\left|}\mathbf{q}\mathbf{\right|}}{{\mathbf{\tau \tau \epsilon }}_{\mathbf{0}}{\mathbf{r}}^{\mathbf{2}}}\mathbf{,}$

Where |q| is the magnitude of the charge on the sphere.

Formula:

$\frac{\mathbf{e}\mathbf{\left|}\mathbf{q}\mathbf{\right|}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}{\mathbf{r}}^{\mathbf{2}}}\mathbf{=}\frac{{\mathbf{mv}}^{\mathbf{2}}}{\mathbf{r}}$

he force must be inward toward the center of the sphere, and since the proton is positively charged, the electric field must also be inward.

$$\begin{gathered} \left|\mathrm{q}\right|=\frac{{\mathrm{mv}}^{2}4{\mathrm{\pi \epsilon }}_0\mathrm{r}}{\mathrm{e}} \\ =\frac{\left(1.67\times {10}^{-27}\mathrm{kg}\right){\left(3.0\times {10}^5\mathrm{m}/\mathrm{s}\right)}^2\left(0.01\mathrm{m}\right)}{\left(9\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\right)\left(1.6\times {10}^{-19}\mathrm{C}\right)} \\ =1.04\times {10}^{-9}\mathrm{C} \end{gathered}$$

The charge on the sphere is negative: