Question

what is the potential energy of of a charge of 2.5*10^-6 C at a point 150 mm from the centre of sphere?

Answer 1

What is the charge distribution like? Is the charge at the center? What is there sphere there for?

Answer 2

there is a single chare which is in the sphere.

Answer 3

use gauss's law to find E, then use this E to find V
\[\iint\limits_{Closed\:Surface} E \:dA=\frac{Q_{enclosed}}{\epsilon_{0}}\]
\[E=-\frac{dV}{dr}\]
Since we're dealing with a sphere, it's symmetrical. So E is the same throughout and is a constant
\[E\iint\limits_{Closed\:Surface} \:dA=E4\pi r^{3} = \frac{Q_{enclosed}}{\epsilon_{0}}\]
And thus
\[E=\frac{Q}{4\pi \epsilon_{0} r^2}\]
Since \[E=-\frac{dV}{dr}\]
\[Edr=-dV\]
and thus
\[\int_{\infty}^{r} Edr=\int_{0}^{V} -dV\]
\[\int_{\infty}^{r}\frac{Q}{4\pi \epsilon_{0} r^2}dr=-(V-0)\]
\[-\frac{Q}{4\pi \epsilon_{0} r}+0=-V\]
\[V= \frac{Q}{4\pi \epsilon_{0} r}\]

Plug in the relevant values of Q and r and you are done!

Answer 4

by putting values in given eq. i got wrong answer

Answer 5

the answer in my book is 6.375*10^-4 J

Answer 6

So what is the charge of the other particle? Since it's potential energy, it must be between 2 charges