A thin glass rod is bent into a semicircle of radius R.A charge + Q is uniformly distributed along the upper half,and a charge -Q is uniformly distributed along the lower half,as shown in the figure.If Q = 10 nC and the electric field E at point P is 5.85 * 10⁴ N/C in -y direction,what is the radius of the semicircle,R? Point P is at the center of the semicircle.

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The Answer of A thin glass rod is bent into a semicircle of radius R.A charge + Q is uniformly distributed along the upper half,and a charge -Q is uniformly distributed along the lower half,as shown in the figure.If Q = 10 nC and the electric field E at point P is 5.85 * 10⁴ N/C in -y direction,what is the radius of the semicircle,R? Point P is at the center of the semicircle.

A Thin Glass Rod Is Bent into a Semicircle of Radius