A plastic disk of radius R = 64.0 cm is charged on one side with a u

asked 2020-11-11

A plastic disk of radius R = 64.0 cm is charged on one side with a uniform surface charge density \(\displaystyle\sigma={6.28}\ {f}\frac{{C}}{{m}^{{2}}}\), and then three quadrants of the disk are removed. The remaining quadrant is shown in Figure. With V = 0 at infinity, what is the potential dueto the remaining quadrant at point P, which is on thecentral axis of the original disk at a distance D = 25.9cm from the original center?

Answers (1)

Total potential at P due to whole
\(\displaystyle={6.28}\cdot{10}^{{-{15}}}\ \frac{{C}}{{m}^{{2}}}\sigma_{{0}}\)
After getting V POTENTIAL DUE TO ONE QUARTER \(\displaystyle={\frac{{{V}}}{{{4}}}}\)
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