my answers are incorrect. i think i am having trouble with not knowing which uni
ID: 1301299 • Letter: M
Question
my answers are incorrect. i think i am having trouble with not knowing which unit to use when converting...
=((9*10^9)*(5.55*10^-11))/(0.0259) = 19.2857143
=19.2857-9.23 = 10.0557143
=((9*10^9)*(5.55*10^-11)) / 10.0557 = 0.04967
For a single, isolated point charge carrying a charge of q = 5.77 times 10 -11 C, one equipotential surface consists of a sphere of radius 0.0259 m centered on the point charge as shown. What is the potential on this surface? You would like to draw an additional equipotential surface, which is separated by 9.23 V from the previously mentioned surface. How far from the point charge should this surface be? This surface must also meet the condition of being farther from the point charge than the original equipotential surface.Explanation / Answer
a. electric potential due to point charge q is V = Kq/r
so V = 9e9 * 5.77 *10^-11/0.0259
V = 20.05 volts
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b. Vnet = V1-V2
V2 = 20.09-9.23
V2 = kQ/R = 10.86
R = 9e9 * 5.77*10^-11/(10.86)
R = 0.0478 m rom the centre of the sphere
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