{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

w13_solution

# Its a weird looking coordinate system but it makes

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: s the integral a bit easier to evaluate and gives the same answer.) A) B) C) D) E) kq L / 2 dx L / 2 d L / 2 x L 2kq L / 2 dx L 0 d x kq L / 2 dx L L / 2 x d kq L dx 0 d L / 2 x L kq L dx L 0 x d V V V V V Version 1 Page 8 13. Charge is distributed on the surface of two very large insulating slabs such that σ1=σ3=σ4=σ and σ2=-σ. What is the electric field at point C? To keep proper track of signs I follow this procedure: 1) draw E-field arrow for each sheet at point C without considering the sign of the sheet (I treat them all positive). 2) write an equation for E at point C. Arrows that point left get a negative sign. 3) plug in the actual value of charge density for each sheet (including the sign this time). This way I count sign from charge only once. A) 0 B) E 2 0 C) 3 E 2 0 D) 3 E 2 0 E) E 2 0 Version 1 Page 9 14. Approximately what is the electric field at point P given that x>>d? The x components cancel. We need to find the y component. Due to symmetry we can find y component of field from posit...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online