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

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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)

2020-11-12
Total potential at P due to whole
$$\displaystyle{V}={\left(\frac{\sigma}{{2}}\sigma_{{0}}\right)}{\left(\sigma{R}^{{2}}+{D}^{{2}}\right)}-{D}\sigma$$
$$\displaystyle={6.28}{f}\frac{{C}}{{m}^{{2}}}$$
$$\displaystyle={6.28}\cdot{10}^{{-{15}}}\ \frac{{C}}{{m}^{{2}}}\sigma_{{0}}$$
$$\displaystyle={8.85}\cdot{10}^{{-{12}}}{\frac{{{C}^{{2}}}}{{{N}{m}^{{2}}}}}$$
$$\displaystyle{R}={64}{c}{m}={0.64}{m}$$
$$\displaystyle{D}={25.9}{c}{m}={0.259}{m}$$
PLUG THE VALUES WE GET V
After getting V POTENTIAL DUE TO ONE QUARTER $$\displaystyle={\frac{{{V}}}{{{4}}}}$$
$$\displaystyle={1.357}\cdot{10}^{{-{5}}}$$

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