1.A charge of -5.40 C is located on the x axis at x = 2.40 cm. A charge of -5.40
ID: 1398751 • Letter: 1
Question
1.A charge of -5.40 C is located on the x axis at x = 2.40 cm. A charge of -5.40 C is located on the x axis at x = -2.40 cm.
Find the electric field on the y axis at y = 6.50 cm. Give only the magnitude of the field.
incorrect answers are
2.Find the force on a 1.43 nC charge on the y axis at y = 6.50 cm. Give only the magnitude of the force.
answer must be four number or more .
2 Incorrect. (Try 1) Sat Jul 11 05:23:58 pm 2015 (EDT) 10.12*10^6 N/C 3 Units required. 6.851*10^6 4 Incorrect. (Try 2) Sun Jul 12 11:10:00 pm 2015 (EDT) 6.851*10^6 N/C 5 Units required. 10.207 * 10^6 6 Submission not graded. Use fewer digits. 10.207 * 10^6 N/C 7 Incorrect. (Try 3) Tue Jul 14 12:44:28 am 2015 (EDT) 10.21 * 10^6 N/C 8 Incompatible units. No conversion found between "N" and the required units. 0.027 N 9 Units required. 1.012 * 10^7 10 Incorrect. (Try 4) Tue Jul 14 11:36:50 am 2015 (EDT) 1.012 * 10^7 N/C 11 Submission not graded. Use more digits. 1.9* 10^7 N/C 12 Incorrect. (Try 5) Tue Jul 14 12:11:21 pm 2015 (EDT) 1.69* 10^7 N/CExplanation / Answer
electric field is same at the y = 6.5 cm due to both the charges
required Electric field is Enet = 2*E*sin(theta)
E = k*q/r^2 = (9*10^9*5.4*10^-6)/((2.4^2+6.5^2)*10^-4) = 10.12*10^6 N/C
Sin(theta) = 6.5/(sqrt(6.5^2+2.4^2)) = 0.938
then Enet = 2*10.12*10^6*0.938 = 18.98*10^6 N/C
2) Force F = q*E = 1.43*10^-9*18.98*10^6 = 0.0271414 N
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