A particle of charge 2.0 times 10^-8 C experiences an upward force of magnitude

Question

A particle of charge 2.0 times 10^-8 C experiences an upward force of magnitude 4.0 times 10^-6 N when it is placed in a particular point in an electric field. What is the electric field at that point? Select the correct answer 1.0 times 10^-2 N/C 8.0 times 10^-2 N/C 2.0 times 10^2 N/C 8.0 times 10^3 N/C 2.0 times 10^4 N/C Point charges q_1 = 50 mu C and q_2 = 25 mu C are placed 1.0 m apart. What is the electric field at a point midway between them? Select the correct answer E = 2.00 times 10^6 N/C towards q_1 E = 2.70 times 10^6 N/C towards q_2 E = 4.80 times 10^6 N/C towards q_2 E = 1.75 times 100 N/C towards q_2 E = 2.50 times 10 N/C towards q_2

Explanation / Answer

Answer(of Question 2)

. As per the Electrostatistics law

F = QE

4x10^-6 = 2x10^-8 E

E = 2x10^2 N/C

(ii)

Answer (of Question3)

E = 1*Q / 4(R)

Net electric field at mid point equals

= 2.50 x10^5 N/C

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