- #1
UnD3R0aTh
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1. The problem is in the attachments.
2. Field strength is equal to voltage divided by distance
3. I want to make sure that i understand a few things before i solve this problem! i would like to know your opinion!
a) first, field strength is inversely proportional to the the distance squared that means filed strength decreases with distance, and as the probe approaches the planet that must mean that the charge is not coming from the planet, it's coming from behind the probe!
b) the negative sign here means that the charge is negative
c) I'm not really sure about this so please could u shed some light on it, i want to know what happens to the voltage as the distance increases from the charge, ite decreases correct? and what would the negative charge of the voltage difference mean?! excess in negative charge of the point closer to the charge ?!
d) my attempt at the solution, the field strength has to be uniform so i take the average which is 500+400/2= -450, then v = field strength times distance which is 20,000,000 meters = -9000MV
2. Field strength is equal to voltage divided by distance
3. I want to make sure that i understand a few things before i solve this problem! i would like to know your opinion!
a) first, field strength is inversely proportional to the the distance squared that means filed strength decreases with distance, and as the probe approaches the planet that must mean that the charge is not coming from the planet, it's coming from behind the probe!
b) the negative sign here means that the charge is negative
c) I'm not really sure about this so please could u shed some light on it, i want to know what happens to the voltage as the distance increases from the charge, ite decreases correct? and what would the negative charge of the voltage difference mean?! excess in negative charge of the point closer to the charge ?!
d) my attempt at the solution, the field strength has to be uniform so i take the average which is 500+400/2= -450, then v = field strength times distance which is 20,000,000 meters = -9000MV
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