Q1) (25p) Two point charges q1= +5µC and q2= -2µC are located as shown in Fig. Find at a point P(3,4) , (a)(15p) the electric field vector and its magnitude (b) (10p) the electric potential. (k = 9x10^9 N.m^2 / C^2)
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Let's call the distance between P and q1 as r1 and the distance between P and q2 as r2. Using the distance formula, we have: r1 = sqrt((3-0)^2 + (4-0)^2) = sqrt(9 + 16) = 5 r2 = sqrt((3-6)^2 + (4-0)^2) = sqrt(9 + 16) = 5 Now, we can find the electric field Show more…
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