Saturday

September 20, 2014

September 20, 2014

Posted by **habtamu** on Saturday, August 11, 2012 at 7:35am.

- physics -
**drwls**, Saturday, August 11, 2012 at 10:39amCharge per unit length along the semicircle is Q/(pi*R). Net field E will be perpendicular to the diameter that runs from one end to the other. Use Coulomb's Law and integrate along the semicircle.

E = [kQ/(pi*R)]*[R*d(theta)/R^2]*

Integral of cos theta from -pi/2 to pi/2.

k is the Coulomb constant.

I get E = 2 k Q/(pi*R^2)

**Answer this Question**

**Related Questions**

Physics Help - Charge Q is uniformly distributed around the perimeter of a ...

physics - A charge of 21 nC is uniformly distributed along a straight rod of ...

physics - Consider a thin spherical shell of radius 16.0 cm with a total charge ...

physics - A point charge of 2.4 μC is located at the center of a spherical ...

Physics - Two spherical shells have a common center. A -3.00 × 10-6 C charge is ...

Physics - Two spherical shells have a common center. A -3.00 × 10-6 C charge is ...

Physics - Two spherical shells have a common center. A -3.00 × 10-6 C charge is ...

physics - A thin ring of radius equal to 25 cm carries a uniformly distributed ...

Physics - A thin ring of radius equal to 25 cm carries a uniformly distributed ...

physics - A plastic rod has been bent into a circle of radius R = 8.80 cm. It ...