Two point charges q 1 = +2.90 nC and q 2 = -7.30 nC are 0.100 m apart. Point A i

Question

Two point charges q1 = +2.90 nC and q2 = -7.30 nC are 0.100 m apart. Point A is midway between them and point B is 0.080 m from q1 and 0.060 m from q2. Take the electric potential to be zero at infinity.

Two point charges q1 = +2.90 nC and 92 =-7.30 nC are 0.100 m apart. Point A is midway between them and point B is 0.080 m from q1 and 0.060 m from 92, Take the electric potential to be zero at infinity 0.050 m 0.050 m (a) Find the potential at point A (b) Find the potential at point B c) Find the work done by the electric field on a charge of 2.50 nC that travels from point B to point A

Explanation / Answer

(a) potential at A

VA = kq1/r1 + kq2/r2
k = 1/4o = 8.987 x 109 N m2/C2.
q1 = 2.9 x 10-9 C, q2 = - 7.3 x 10-9 C
r1 = r2 = 0.5 m

VA = [(8.987 x 109) x (2.9 x 10-9) / 0.05 - (8.987 x 109) x (7.3 x 10-9) / 0.05]
= - 790.856 V

(b) Potential at point B
VB = kq1/r1 + kq2/r2
r1 = 0.08 m, r2 = 0.06 m
VB = [(8.987 x 109) x (2.9 x 10-9) / 0.08 - (8.987 x 109) x (7.3 x 10-9) / 0.06]
- 767.64 V

(c)
Work done in moving the charge q from B to A
W = q x [Change in potential]
= q x [VB - VA]
= 2.5 x 10-9 x [- 790.856 - ( - 767.64)]
= - 58.04 x 10-9 J

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