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two charges 3.8 into 10 to the power 5 cm and 5.5 into 10 power 4 Siya place to get the distance and 10 cm each and other find the the value of electricity force acting between them


Navnath Maynale 24th Aug, 2020
Answer (1)
Prabhav Sharma 24th Aug, 2020

Hello,

Here, Q1 = 3.8 * 10^5 and Q2 = 5.5 * 10^4

Distance between them, r = 10*10^-2

According to Coloumb's law, force acting between two bodies having charged Q1 and Q2 placed r distance apart is given by:

F = 1/4(pie)(Epsilon) * (Q1*Q2)/r^2

Substituting the values in the above equation, we get:

F = (1/4 * 3.14 * 8.85 * 10^-12) * (3.8 * 10^5 * 5.5 * 10^4) / (10*10^-2)^2

F = 1.88 * 10^22 N (Answer)

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