Question

moment that is perpendicular toa radial line and has a magnitude of 1.28 × 10-21 C-m What is the net electric potential at the center? 31 SSM WWW A plastic disk of radius R 64.0 cm is charged on one side with a uniform surface charge density σ=7.73 fC/m2, and then three quadrants of the disk are removed.The remaining quadrant is shown in Fig. 24-50. With V 0 at infinity, what is the potential due to the remaining quad- rant at point P, which is on the central axis of the original disk at distance D 25.9 cm from the original center? Figure 24-50 Problem 31.

For problem 24.31 of the text, calculate the potential at point P in mV for the case that the removed part of the disk subtends an angle of 291

Answer 1

answer) it is given in the question that the disk is uniformly charged, now this means that the potential at point P , which is a single quadrant is one-fourth of the potential of the full disk

potential for full disk is given by

V=$\sigma$/2$\epsilon$_o*(( $\sqrt{}$R²+D²)-D)

so for single quadrant , potential V1=V/4

so V1=$\sigma$/8$\epsilon$_o(($\sqrt{}$R²+D²)-D=(7.73*10^-15C/m²/8*8.85*10^-12C²/N.m²)*$\sqrt{}$0.64²+0.259²)-0.259=0.0471mV

so the answer is 0.047mV or 0.0471mV or 0.05mV. or 4.71*10^-5V