What are the magnitude and direction of the electric field at a distance of 1.50

Question

What are the magnitude and direction of the electric field at a distance of 1.50 m from a 50.0-nC charge? 10 N/C away from the charge 20 N/C away from the charge 20 N/C toward the charge 200 N/C toward the charge 200 N/C away from the charge A force of 10 N acts on a charge of 5.0 mu C when it is placed in a uniform electric field. What is the magnitude of this electric field? 1.0 MN/C 1000 MN/C 50 MN/C 0.50 MN/C 2.0 MN/C Two point charges of +20.0 mu C and -8.00 mu C are separated by a distance by a distance of 20.0 cm. What is the intensity of electric field E midway between these two charges? 25.2 times 10^4 N/C directed towards the negative charge 25.2 times 10^5 N/C directed towards the negative charge 25.2 times 10^5 N/C directed towards the positive charge 25.2 times 10^6 N/C directed towards the negative charge

Explanation / Answer

Question 5 )

Electric Force is given by the formula

F = q*E

Here E is the electric field and q is the charge

Given , F = 10 N and charge = 5*10^-6 C

E = F/q

= 10/5*10^-6

= 2*10^6 N/C = 2 MN/C

Therefore) answer ) option E) 2 MN/C

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