Explain the differences between insulators, and conductors. How can a charged bo

Question

Explain the differences between insulators, and conductors. How can a charged body be attracted to an uncharged body? Explain polarization and induction? The figure below shows three small charges A, B, P in a line The charge at A is positive, that at B negative, and that at P is positive. The values are those shown. Calculate the force on the charge at P due to A and B. At what point x on the line AB could there be no force on the charge P due to A and B P were placed there? In the figure below, two small equal charges 2 times 10^-8 C are placed at A and B, one positive other negative. AB is 6cm. Find the force on a charge +1 times 10^-8 C placed at P, where P is 4 from the line AB along the perpendicular bisector XP The work per coulomb between two points A and B which is the work an external to carry a unit positive charge from B to A is the potential difference V_AB = kQ(l/a - 1/b) point charges 12 and 8 micro coulomb respectively are 10 cm apart find the work in closer.

Explanation / Answer

Question 4) Qa = 2*10^-8 C, Qb = 5*10^-8 C, Qp = 1*10^-8 C

Force on P due to A is F1 = kQa*Qp/r2

r = 0.06 + 0.04 = 0.1 m

Hence F1 = 9*109 * 2*10^-8 * 1*10^-8 / 0.12 = 1.8 * 10-4 N (toward right)

Similarly force due to Qb on P

F2 = kQb*Qp/r2, r = 0.04

F2 = 9*109 * 5*10^-8 * 1*10^-8 / 0.042 = 2.8 * 10-3 N (towards left)

Net force |Fnet| = F1 - F2 = 1.8 * 10-4 - 2.8 * 10-3 = 2.63 * 10-3 (towards left)

b) The net force due to a and b can be zero only on left of Qa because in the middle both points in same direction and on the right of Qb, F2 is always greater than F1

Hence let Fnet is zero at distance x from Qa towards left

then |F1| = |F2|

kQaQp/x2 = kQbQp/(0.06+x)2

2/x2 = 5/(0.06+x)2

(0.06+x)/x = (5/2)1/2

0.06/x = 0.58

x = 0.103 m

Related Questions

Navigate

• Browse (All)
• Browse E
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.