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Problem 19.31 Problem 19.31 Part A Find the direction and magnitude of the net e

ID: 2280871 • Letter: P

Question

Problem 19.31 Problem 19.31 Part A Find the direction and magnitude of the net electrostatic force exerted on the point charge q2 in the figure.(Figure 1) Let q=+2.5?C and d=33cm. Express your answer using two significant figures. Fnet =
N SubmitMy AnswersGive Up Incorrect; Try Again; 5 attempts remaining Part B Express your answer using three significant figures. ? =
? counterclockwise from q2-q3 direction SubmitMy AnswersGive Up Incorrect; Try Again; 5 attempts remaining Provide Feedback Part A Find the direction and magnitude of the net electrostatic force exerted on the point charge q2 in the figure.(Figure 1) Let q=+2.5?C and d=33cm. Express your answer using two significant figures. Fnet =
N SubmitMy AnswersGive Up Incorrect; Try Again; 5 attempts remaining Part B Express your answer using three significant figures. ? =
? counterclockwise from q2-q3 direction SubmitMy AnswersGive Up Incorrect; Try Again; 5 attempts remaining Part A Find the direction and magnitude of the net electrostatic force exerted on the point charge q2 in the figure.(Figure 1) Let q=+2.5?C and d=33cm. Express your answer using two significant figures. Fnet =
N SubmitMy AnswersGive Up Incorrect; Try Again; 5 attempts remaining Part B Express your answer using three significant figures. ? =
? counterclockwise from q2-q3 direction SubmitMy AnswersGive Up Incorrect; Try Again; 5 attempts remaining Part A Find the direction and magnitude of the net electrostatic force exerted on the point charge q2 in the figure.(Figure 1) Let q=+2.5?C and d=33cm. Express your answer using two significant figures. Fnet =
N SubmitMy AnswersGive Up Incorrect; Try Again; 5 attempts remaining Part B Express your answer using three significant figures. ? =
? counterclockwise from q2-q3 direction SubmitMy AnswersGive Up Incorrect; Try Again; 5 attempts remaining Part A Find the direction and magnitude of the net electrostatic force exerted on the point charge q2 in the figure.(Figure 1) Let q=+2.5?C and d=33cm. Express your answer using two significant figures. Fnet =
N SubmitMy AnswersGive Up Incorrect; Try Again; 5 attempts remaining Part A Find the direction and magnitude of the net electrostatic force exerted on the point charge q2 in the figure.(Figure 1) Let q=+2.5?C and d=33cm. Express your answer using two significant figures. Fnet =
N SubmitMy AnswersGive Up Incorrect; Try Again; 5 attempts remaining Fnet =
N Fnet =
N SubmitMy AnswersGive Up Incorrect; Try Again; 5 attempts remaining Incorrect; Try Again; 5 attempts remaining Incorrect; Try Again; 5 attempts remaining Part B Express your answer using three significant figures. ? =
? counterclockwise from q2-q3 direction SubmitMy AnswersGive Up Incorrect; Try Again; 5 attempts remaining Part B Express your answer using three significant figures. ? =
? counterclockwise from q2-q3 direction SubmitMy AnswersGive Up Incorrect; Try Again; 5 attempts remaining ? =
? counterclockwise from q2-q3 direction ? =
? counterclockwise from q2-q3 direction SubmitMy AnswersGive Up Incorrect; Try Again; 5 attempts remaining Incorrect; Try Again; 5 attempts remaining Incorrect; Try Again; 5 attempts remaining Provide Feedback of 1 Problem 19.31 Part A Find the direction and magnitude of the net electrostatic force exerted on the point charge q2 in the figure.(Figure 1) Let q=+2.5?C and d=33cm. Express your answer using two significant figures. Fnet =
N SubmitMy AnswersGive Up Incorrect; Try Again; 5 attempts remaining Part B Express your answer using three significant figures. ? =
? counterclockwise from q2-q3 direction SubmitMy AnswersGive Up Incorrect; Try Again; 5 attempts remaining Provide Feedback of 1

Explanation / Answer

l

q = 2.7*10^-6 C

q1 = q

q2 = -2q

q3 = -3q

q4 = -4q

d = 33 cm = 0.33 m

Force on q2 due to q1, F1 = [Kq1q2/d^2](-j)

                                      = [2Kq^2/d^2](-j)

                                      = - 1.204 j N

Force on q2 due to q3, F3 = [Kq2q3/d^2](i)

                                      = [6Kq^2/d^2](i)

                                      = 3.614 i N

Force on q2 due to q4, F4 = [Kq2q4/(?2 d)^2][cos45 i - sin45 j]

                                      = [8Kq^2/(?2 d)^2][0.707 i - 0.707 j]

                                      = 2.408*0.707[i - j] N

                                      = 1.702 [ i - j] N

Net force on q2, F = F1 + F3 + F4

                           = -1.204 j + 3.614 i + 1.702 i - 1.702 j

                           = 5.316 i - 2.906 j

Magnitude : |F| = 6.058 N

Direction ? = Tan^-1[2.906/5.316]

                 = 28.66 degrees below the X axis.