# Question #24b79

Jun 25, 2016

$= \frac{Q}{8 \pi {\epsilon}_{o} {a}^{2}}$

$= \frac{k Q}{2 {a}^{2}}$

#### Explanation:

in spherical coordinates, using this coordinate system

we arrange the axes so that cup in question is symmetrically placed about origin O with the centre of the flat surface at O, and we place a unit test charge ${Q}_{t}$ at O to calculate the field there.

The surface element for constant radius a follows as
$\mathrm{dS} = {a}^{2} \setminus \sin \theta \setminus d \setminus \theta \setminus d \setminus \varphi$

Assuming a uniform charge density $\sigma$ on the cup, the charge of a small element is

$\mathrm{dq} = \sigma \mathrm{dS}$ where $\sigma = \frac{Q}{A} = \frac{Q}{2 \pi {a}^{2}}$ ,

the force due to a given small element will follow Couloumb's Law: $F = k \frac{Q {Q}_{t}}{r} ^ 2$

$\mathrm{dF} = k \frac{\mathrm{dq} {Q}_{t}}{a} ^ 2$

the force on the unit test charge ${Q}_{t}$ placed at O will be zero due to symmetry. we need only look at the force in the z-direction. so here we can say that

${\mathrm{dF}}_{z} = k \frac{\mathrm{dq} {Q}_{t}}{a} ^ 2 \cdot \cos \theta$

Putting it altogether

${\mathrm{dF}}_{z} = k \frac{\left(\sigma \mathrm{dS}\right) {Q}_{t}}{a} ^ 2 \cdot \cos \theta$

$\frac{{\mathrm{dF}}_{z}}{Q} _ t = {\mathrm{dE}}_{z} = k \frac{\sigma}{a} ^ 2 \setminus \cos \theta \setminus \mathrm{dS}$

$\setminus \implies {\mathrm{dE}}_{z} = k \frac{\sigma}{a} ^ 2 \setminus \cos \theta \setminus {a}^{2} \setminus \sin \theta \setminus d \setminus \theta \setminus d \setminus \varphi$

$= k \frac{\sigma}{2} \setminus \sin 2 \theta \setminus d \setminus \theta \setminus d \setminus \varphi$

and as $k = \frac{1}{4 \pi {\epsilon}_{o}}$

$\setminus \implies {\mathrm{dE}}_{z} = \frac{1}{4 \pi {\epsilon}_{o}} \frac{Q}{4 \pi {a}^{2}} \setminus \sin 2 \theta \setminus d \setminus \theta \setminus d \setminus \varphi$

$\setminus \implies {\mathrm{dE}}_{z} = \frac{1}{{\left(4 \pi a\right)}^{2} {\epsilon}_{o}} Q \setminus \sin 2 \theta \setminus d \setminus \theta \setminus d \setminus \varphi$

$\setminus \implies {E}_{z} = \frac{1}{{\left(4 \pi a\right)}^{2} {\epsilon}_{o}} Q {\int}_{\varphi = 0}^{2 \pi} \setminus {\int}_{\theta = 0}^{\frac{\pi}{2}} \setminus \setminus \sin 2 \theta \setminus d \setminus \theta \setminus d \setminus \varphi$

$= 2 \pi \frac{1}{{\left(4 \pi a\right)}^{2} {\epsilon}_{o}} Q \setminus {\int}_{\theta = 0}^{\frac{\pi}{2}} \setminus \setminus \sin 2 \theta \setminus d \setminus \theta$

$= \frac{1}{8 \pi {\epsilon}_{o} {a}^{2}} Q {\left[- \frac{1}{2} \setminus \cos 2 \theta\right]}_{\theta = 0}^{\frac{\pi}{2}} \setminus$

$= \frac{1}{8 \pi {\epsilon}_{o} {a}^{2}} Q {\left[\setminus \cos 2 \theta\right]}_{\theta = \frac{\pi}{2}}^{0} \setminus$

$= \frac{Q}{8 \pi {\epsilon}_{o} {a}^{2}}$

$= \frac{k Q}{2 {a}^{2}}$