7. 1/3 points I Previous Answers SerCP11 15.P.018. My Notes Ask Your Teacher (a)

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7. 1/3 points I Previous Answers SerCP11 15.P.018. My Notes Ask Your Teacher (a) Determine the electric field strength at a point 1.00 cm to the left of the middle charge shown in the figure below. Enter the magnitude of the electric field only.) N/C 6.00 C 1.50 AC -200 C -3.00 cm-+-2.00 cm (b) If a charge of-2.40 c is placed at this point, what are the magnitude and direction of the force on it? magnitude direction toward the left Need Help? Read It Submit Answer Save ProgressPractice Another Version

Explanation / Answer

The electric field at point 1 cm to the left will be the sum of individual fields doe to three charges.

R = E1 + E3 - E2

since field is directed away from positive charge and towards the negative.

E = kq1/r1^2 + kq3/r3^2 - kq2/r2^2

E = 9 x 10^9 x 10^-6 (6/0.02^2 + 2/0.03^2 - 1.5/0.01^2) = 2 x 10^7 N/C

Hence, E = 2 x 10^7 N/C

b)F = qE

F = -2.4 x 10^-6 x 2 x 10^7 = -48 N

Hence, F = 48 N ; towards the left

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