Problem 11 (6.25 pts): A dipole has charges 3 × 10-5 C and-3 × 10-5 C. Along the

Question

Problem 11 (6.25 pts): A dipole has charges 3 × 10-5 C and-3 × 10-5 C. Along the bisector of this dipole, 0.8 m from its center, you measure an electric field of magnitude 1580 N/C. What is the charge separation between the two point charges? 5.98 × 10-3 C B)2.99 × 10-3 C 3.75 × 10-3 C D) 5.46 × 10-2 c E) 1.53 x 10-3 C Problem 12 (6.25 pts): When two point charges are 2.0 cm apart, each one experiences a 1.0-N electric force due to the other charge. If they are moved to a new separation of 8.0 cm, the electric force on each of them is closest to A) 1 N B) 16 N C) 0.25 N 4.0 N E) 0.063 N

Explanation / Answer

Problem 11)

Field on the bisector of the dipole is E = k*q*d/(r^2+(d/2)^2)^(3/2)

given that q = 3*10^-5 C

d = ? m

r = 0.8 m

then E = k*q*d/((r^2+(d/2)^2)^(3/2)

1580 = (9*10^9*3*10^-5*d)/(0.8^2+(d/2)^2)^(3/2)

1580*(0.8^2+(d/2)^2)^(3/2) = (9*10^9*3*10^-5*d)

solving we get d = 2.99*10^-3 m

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