Question

Solve part b: Note the answer should be a vector

(1 point) Sketch S, the part of the plane - 10x + 5y + 10z = -39 selected by the inequality x2 + y2 + z2 <9. Assume a uniform surface charge with density ps = 36 nC/m² on S, with no other charges anywhere in space. Both parts below refer to the resulting electric field at the origin, E(O). (a) Find the point P at which E(P) = -E(O). ANSWER: P= (52/15,-26/15,-52/15) (b) Find E(O). ANSWER: E(O) = N/C. Hint: Draw a picture. Could you do the problem if the given plane were simply z = 2.6?

Answer 1

Electric field due to surface charge, E =  
LON 250

ps = Surface charge density

E=
36 + 10-4 2* 8.854 * 10-12
= 2.03 * 10³$\widehat{a_{N}}$ N/C.

When the plane is z=2.6, then the normal direction will be a_z. That is $\widehat{a_{N}}$ = $\overrightarrow{a_{z}}$

E= 2.03 * 10³ $\overrightarrow{a_{z}}$ N/C

Since origin is below z=2.6, the electric field at origin E(0) will be negative of the net electric field due to given surface charge.

E(0) =  - 2.03 * 10³ $\overrightarrow{a_{z}}$ N/C