First, let us find the magnitude of the electric field at P due to each charge.
ID: 1573916 • Letter: F
Question
First, let us find the magnitude of the electric field at P due to each charge. The fields E1 due to the 7.0 C charge and E2 due to the-70 charge are shown in Figure 23.14. Their magnitudes are Ei = k,19T = 8.99 x 109 N·m/C27.0 x 10-6 C (0.40 m) = 393312.5 N/C 7.0 × 10-6 C (0.50 m) 92 251720 The vector El has only a y component. The vector E2 has a x component given by E2 cos = (3/5E2 and a negative y component given by-E2 sin =-(4/5JE2. Hence, we can express the vectors as J N/C E2= ( The resultant field E at P is the superposition of E1 and E2: i) N/c E=E1+ E2 = j) N/c Exercise 23.5 Hints: Getting Started | I'm Stuck Suppose the sign of charge q2 is changed to a positive +7.0 C. Determine the electric field at point P in this case. j) N/CExplanation / Answer
E1 = 393312.5 j
E2 = (3/5) (251720) i + (-4/5) (251720) j = (151032) i + (- 201376) j
The resultant electric field is given as
E = E1 + E2 = 393312.5 j + (151032) i + (- 201376) j
E = (151032) i + (- 191936.5) j
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