Suppose a charge of 2.30 µC is at the origin and a charge of 3.10 µC is at the p

Question

Suppose a charge of 2.30 µC is at the origin and a charge of 3.10 µC is at the point (0, 3.00) m.

(a) Find the electric potential at (4.00, 0) m, assuming the electric potential is zero at infinity.
___ V

(b) Find the work necessary to bring a 4.10 µC charge from infinity to the point (4.00, 0) m.
___ J

A 4.90-µC point charge is at the origin, and a point charge q2 = 2.30 µC is on the x-axis at (3.00, 0) m, as shown in the figure.

(a) If the electric potential is taken to be zero at infinity, find the electric potential due to these charges at point P with coordinates (0, 4.00) m.
___ V

(b) How much work is required to bring a third point charge of 4.10 µC from infinity to P?
____ J

Explanation / Answer

Given:

Charge at origin, q = -2.3x10-6 C

Charge at (0, 3), Q = 3.1x10-6 C

To Find: (a) Electric potential at A (4, 0), VA; (b) Work to bring charge Q' (4.10x10-6)C to point A, WQ'

Solution:

Distance between charge q and point A, rq = 4 m

Distance between charge Q and point A, rQ = (42 + 32)1/2 = 251/2 = 5 m

Potential at A, VA = kq / rq + kQ / rQ

VA = 9x109 * (-2.3x10-6) / 4 + 9x109 * 3.1x10-6 / 5

VA = -5.175x103 + 5.58x103 = 405 J / C

Ans. Potential at 4,0 due to charges is equal to 405 J / C

(b)

Work done to bring charge Q' to point A from infinity will be

WQ' = -Q' ( VA - Vinfinity)

WQ' = -4.1x10-6 * 405 = -1.6605x10-3 J

Ans. Work done to bring the charge is 1.6605x10-3 J

Given:

Charge at origin, q = 4.9x10-6 C

Charge at (3, 0), Q = -2.3x10-6 C

To Find: (a) Electric potential at P (0, 4), VA; (b) Work to bring charge Q' (4.10x10-6)C to point P, WQ'

Solution:

Distance between charge q and point P, rq = 4 m

Distance between charge Q and point P, rQ = (42 + 32)1/2 = 251/2 = 5 m

Potential at P, VP = kq / rq + kQ / rQ

VP = 9x109 * (4.9x10-6) / 4 + 9x109 * (-2.3x10-6) / 5

VP = 11.025x103 - 4.14x103 = 6885 J / C

Ans. Potential at P due to charges is equal to 6885 J / C

(b)

Work done to bring charge Q' to point P from infinity will be

WQ' = -Q' ( VP - Vinfinity)

WQ' = -4.1x10-6 * 6885 = -0.02823 J

Ans. Work done to bring the charge to P is 0.02823 J

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