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
Answer (1)
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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