(6%) Problem 17: Two point charges Q1 5.9 mC and Q2 11 mC are fixed on the x-axi

Question

(6%) Problem 17: Two point charges Q1 5.9 mC and Q2 11 mC are fixed on the x-axis at points (td, 0), whcre d 0.042 m, as shown in the figure. Consider the labeled points at (tdr, 0) and (0, +dy), where d 0.019 m and d. 0.029 m. A 25% Part (a) Find the potential at point P1,in gigavolts, due to the two point charges A 25% Part (b) Find the potential at point P2, in gigavolts, due to the two point charges A 25% Part (c) Find the potential at point P in megavolts, due to the two point charges. A 25% Part (d) Find the potential at point P4, in megavolts, due to the two point charges T P Otheexpertta.com

Explanation / Answer

the potential at a point due to a charge A is KQ/R

a.) at P1, due to Q1

R1 = d-dx = 0.023 m

so V1 = KQ/R = 9 x 109 x 5.9 x 10-3 / 0.023 = 2.30 GV

due to Q2, d2 = d + dx = 0.07 m

ao v2 = k Q2/R2 = 9 x 10^9 x -11 x 10^-3 / 0.07 = -1.4142

so, net v = 2.3 -1.4142 = 0.8857 GV

b.)at p2,

r1 = d + dx = 0.07 m

r2 = d-dx = 0.023 m

so, v1 = kq1/r1 = 9 x 10^9 x 5.9 x 10^-3 / 0.07 = 0.758571 GV

v2 = kq2/r2 = 9 x 10 ^9 x -11 x 10-3 /0.023 = -4.304 GV

so, net V = -4.304 + 0.7585=-3.54584 GV

c.) at p3,

both q1 and q2 are equidistant from it and that distance r = sqrt (d2 + dy2)

= 0.051039 m

so potential due to both the charges = k (q1+q2)/r = 9 x 109 x (5.9-11) x 10-3/0.05

= -0.918 GV

d.) due to symmetry at P4, potential will be same as that at P3

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