Question Solved1 Answer Problem 5: Charge Q is uniformly distributed along a thin, flexible rod of length L. The rod is then bent into the semicircle shown in the figure: Center L C a. Find an expression for the electric field at the center of the semicircle. Hint: A small piece of arc length As spans a small angle 10 = As/R, where R is the radius. b. Evaluate the field strength if L = 10 cm and Q = 30 nC.

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Transcribed Image Text: Problem 5: Charge Q is uniformly distributed along a thin, flexible rod of length L. The rod is then bent into the semicircle shown in the figure: Center L C a. Find an expression for the electric field at the center of the semicircle. Hint: A small piece of arc length As spans a small angle 10 = As/R, where R is the radius. b. Evaluate the field strength if L = 10 cm and Q = 30 nC.
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Transcribed Image Text: Problem 5: Charge Q is uniformly distributed along a thin, flexible rod of length L. The rod is then bent into the semicircle shown in the figure: Center L C a. Find an expression for the electric field at the center of the semicircle. Hint: A small piece of arc length As spans a small angle 10 = As/R, where R is the radius. b. Evaluate the field strength if L = 10 cm and Q = 30 nC.
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iiven,charge= Qlength =La) an expression for the electric field vec(E) at she center of the semicirde.consider DA - y axisO B - x axisand a small piece of arc length EF which has /_\S spans and it impose Delta theta angle (which is small) on center.so Delta theta=(Delta S)/(R) where R is radius of semicirde.The magnitide of field af 0dE=(1)/(4piepsilon_(0))(Rd theta xx(Q)/(R^(2)))/(pi(R)/(2))then we getdE=(Q)/(2pi^(2)epsi_(0)R^(2))d thetawe see, in the figure that horizontal component of electric field get cande and the resultant field is along OA and vent ... See the full answer