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The charge for Q3 is not given. I could not find formula in book. So do I take EA total 4.5 X 10^6 N/C. And use it for E? Then set formula like so 4.5 X 10^6= (9.0x10^9)Q/(.30m)^2 and solve for Q to get charge? I need help.
ple 16-9, but calculate the electric field (magnitude and direction) at Q1 due to the Rework Exam other two charges.
EXAMPLE 16-9 E above two point charges. Calculate the total ele field at point A in Fig. 16-28 due to both charges, Q, and APPROACH The calculation is much like that of Example 16-4, exce we are dealing with electric fields instead of force. The electric field at point is the vector sunn of the fields EAI due to Qi , and EA2 due to Q2 , we find magnitude of the field produced by each point charge, then we add thein components to find the total field at point A. SOLUTION The magnitude of the electric field produced at point A by each of the charges Q, and Q, is given by E- kO/r2, so pt now AI 52m30 (9.0 × 10° N-㎡/C)(50 × 10-6 C) 1.25 × 10° N/C, FIGURE 16-28 Calculation of the electric field at point A. Example 16-9 . (0.60 m) AI Ev = (9.0× 10 N-m2/C2)(50× 10-6C) 10°N/C. 5.0 . PROBLEM SOLVING A2 (0.30 m)2 Igore signs of charges and determine direction physically shoneing directions on diagram inThe direction of Ea points from A toward Q, (negative charge) whereas E, points from A away from Q. as shown; so the total electric field at A. Ex.has components EAy = EA2-Bw sin 30°-4.4 × 106 N/C. Thus the magnitude of EA is ㄙㄧ ˇ (1.1アー(4.4ア× 10° N/C-45 × 10° N/C. and its direction is ф (Fig. 16-28) given by tano-EAy/EA, 4.4/1.1-40.
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Answer #1

Question is incomplete.

And your approach is wrong too.

Basically electric field can't be completed until the value of Charge Q3 is known.

If you want, you can compute electric field at Q1 due to charge Q2 only, by the same approach.

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