The class is about Electricity and Magnetism. The website wants exactly the perfect answer and in the same metric system units that are stated on the calculators that are on the picture for the question, if not, it'll mark it wrong.

2L A thin rod of length 24 = 6.3 cm is centered on the x-axis. The rod carries a uniformly distributed charge Q = 0.028 uC. Find the potential V at a point on the y axis a distance 4.7 cm from the center of the rod. V = V abc sin(X 0x10 Submit Reset Reset All .Begin by writing an expression for the infinitesimal element dV of potential due to an infinitesimal charged segment of the rod with charge dq = (Q/21)dx. Then integrate dV to obtain the total potential due to all charged segments of the rod. .The integrals are somewhat challenging. You may use Mathematica or tables if you like. If you attempt the integral by hand, begin with a standard trigonometric substitution x = y tan 0. After simplification, you should arrive at the integral of sec 0, which evaluates to In(sec @ + tan 0).Soin ) P K X O) K 21 K Potential at point P due to an infinitesimal charged Segment of the rod having length 'dx & charge, dy= dx is, 2L dv = dq or, dv = - Q dox 2L Q MR 8 2L Now Potential at point 'P' due to all charged elements of the rod is = fav Q dx 2L -L Q = x 2 dox VX ty 2 inGo [ log [ 2 + 12 + 42 ]V = . 1 Q log L t N L ty ? UAE L' = gx10 x 2X0 028 X 10 - 6 6.3 +w( 6 3 ) + ( 4.72 6. 3X162 = 0.8 log 8. 8079 4. 7 - 0. 8 X 0. 628 Volt V = 0.5024 Volt So, the Potential at point P, which is at a distance 4.7 cm from the center of the rod PS, 0.5024 volts