Explain why the answers to part a and part b are not the...

Explain why the answers to Part A and Part B are not the same. Match the words in the left column to the appropriate blanks in the sentences on the right. ResetHelp closer to inverse squared farther away from 1. The decrease in spacing of the equipotentials as you move Blank the charge is due to the fact that the potential is proportional to the Blank distance, not the distance.target 1 of 2target 2 of 2

Transcript

00:01 Hello students, now in this question they have given a constant charge a which has the charge of plus 7 .60 column and it is set it at the origin.

00:12 And they are asking here three part of questions.

00:16 In that part a has the question of they are asking the radial distance between 500 volt equipotential surface and 1 ,000 volt surface and part b has the question of they are asking the distance between 1 ,000 volt surface and the 1 ,000.

00:30 The 1 ,500 volt surface and we have to explain why this part a and part b answers are not same they are asking now first see the electron potential due to the point charge at a distance are from the charge is given by the equation or formula potential v is equal to k q by here k is the coulum's constant it is 9 in 2 to the power of 9 newton meter square per column square and q is the point charge and r is the distance.

01:09 First here we need to calculate radius r by rearranging this equation we have k q by v that is r is equal to k q by v.

01:19 Q is given as plus 7 .6 micro column and also 7 .60 into 10 to the power of minus 6 column.

01:33 So by substituting this in r formula we will get r is equal to 9 into 10 to the power of 9 newton meter square per column square into 7 .60 into 10 to the power of minus 6 divided by divided by we will get r is equal to 68 .4 into 10 to the power of 3 divided by now using this radii we can calculate or we have here we they have given potential we they have given as 500 volts and v is equal to thousand volts and also they have given here v is equal to 4 ,000 500 we have to substitute these three potential values in this equation we will get for v is equal to 500 volts we have 68 .4 into the power of 3 divided by 500 which is equal to 136 .80 meters and then for thousand volts we have 6068 .4 into the power of 3 divided by which is equal to 68 .40 meters and for b is equal to 1 ,500 volts we have r is equal to 68 .4 into the power of 3 divided by 1 ,500 equal to 45 .6 meters now for part a we have question they have given as we have to find the distance between 500 volt and 1 ,000 volt equipotential surface and this can be given by distance between 500 volts and 1 ,000 volts and this is calculated by subtracting this 500 volts from 1 ,000 volts we will get 136 .80 minus 68 .40 which is equal to finally we will get this as 68 .40 meters...

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