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20 positively charged particles are kept fixed on the x-axis at points X equal to 1 metre, 2 metre , 3 metre ,.,20 metre the first particle has a charge 1C the second 8C 3rd 27C and so on find the magnitude of the electric force acting on a a 1 coulomb charge placed at the origin?
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20 positively charged particles are kept fixed on the x-axis at points...
Problem:

20 positively charged particles are kept fixed on the x-axis at points X equal to 1 metre, 2 metre , 3 metre ,.,20 metre the first particle has a charge 1C the second 8C 3rd 27C and so on find the magnitude of the electric force acting on a a 1 coulomb charge placed at the origin?


Solution:


Step 1: Calculate the electric field due to each charged particle

Electric field due to a point charge is given by:

E = kQ/r^2

where, k = Coulomb's constant = 9x10^9 Nm^2/C^2, Q = charge of the particle, r = distance between the particle and the point where electric field is to be calculated.

Using the above formula, we can calculate the electric field due to each charged particle at the origin:



  • Electric field due to first particle = k(1)/1^2 = 9x10^9 N/C

  • Electric field due to second particle = k(8)/2^2 = 2.25x10^10 N/C

  • Electric field due to third particle = k(27)/3^2 = 3x10^10 N/C

  • Electric field due to fourth particle = k(64)/4^2 = 3.6x10^10 N/C

  • ...

  • Electric field due to twentieth particle = k(1048576)/20^2 = 2.31x10^9 N/C



Step 2: Calculate the net electric field at the origin

Electric field is a vector quantity and it follows vector addition. Therefore, we need to add the electric fields due to each particle vectorially to get the net electric field at the origin:

E_net = E1 + E2 + E3 + ... + E20

E_net = 9x10^9 N/C + 2.25x10^10 N/C + 3x10^10 N/C + 3.6x10^10 N/C + ... + 2.31x10^9 N/C

E_net = 1.63x10^11 N/C


Step 3: Calculate the electric force on a 1C charge placed at the origin

Electric force acting on a charge q in an electric field E is given by:

F = qE

Using the above formula, we can calculate the electric force on a 1C charge placed at the origin:

F = 1C x 1.63x10^11 N/C = 1.63x10^11 N


Step 4: Conclusion

Therefore, the magnitude of the electric force acting on
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20 positively charged particles are kept fixed on the x-axis at points X equal to 1 metre, 2 metre , 3 metre ,.,20 metre the first particle has a charge 1C the second 8C 3rd 27C and so on find the magnitude of the electric force acting on a a 1 coulomb charge placed at the origin?
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20 positively charged particles are kept fixed on the x-axis at points X equal to 1 metre, 2 metre , 3 metre ,.,20 metre the first particle has a charge 1C the second 8C 3rd 27C and so on find the magnitude of the electric force acting on a a 1 coulomb charge placed at the origin? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 20 positively charged particles are kept fixed on the x-axis at points X equal to 1 metre, 2 metre , 3 metre ,.,20 metre the first particle has a charge 1C the second 8C 3rd 27C and so on find the magnitude of the electric force acting on a a 1 coulomb charge placed at the origin? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 20 positively charged particles are kept fixed on the x-axis at points X equal to 1 metre, 2 metre , 3 metre ,.,20 metre the first particle has a charge 1C the second 8C 3rd 27C and so on find the magnitude of the electric force acting on a a 1 coulomb charge placed at the origin?.
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