Using the electric field from a point charge, , we can integrate to get the volt
ID: 2053078 • Letter: U
Question
Using the electric field from a point charge, , we can integrate to get the voltage (potential) difference between two points. It is convenient to choose a reference point at and define . The voltage at a point a distance r from the point charge is then.
Then for any collection of point charges, or for any finite charge distribution the voltage at a point can be obtained by doing a sum or integral. Since voltage is a scalar, a single sum or integral is sufficient and no vector components need be considered.
(a) A single charge -6.0 nC is at the origin. Find the voltage at a point located at x = 0.50 m. V
(b) Find the voltage at a point located at x = 0.30 m and y = 0.40 m. V
(c) A second charge 9.0 nC is placed at x = -1.5 m. What is the voltage at the point located at x = 0.50 m? V
(d) What is the voltage at a point on the y axis at y = 1.9 m due to the two charges? V
Explanation / Answer
(a) A single charge -6.0 nC is at the origin. Find the voltage at a point located at x = 0.50 m. V
(b) Find the voltage at a point located at x = 0.30 m and y = 0.40 m. V
(c) A second charge 9.0 nC is placed at x = -1.5 m. What is the voltage at the point located at x = 0.50 m? V
(d) What is the voltage at a point on the y axis at y = 1.9 m due to the two charges?
(a) V = kQ/r = 9x109 x-6.0x10-9 /0.5 = -108 V
(b) V = kQ/r = 9x109 x-6.0x10-9 /0.3 = -180 V
V3 = kQ/r = 9x109 x-6.0x10-9 /0.4 = -135 V
(c) Charge is at x = -1.5 m; Point P is at x = 0.5 m. So, r = 1.5 + 0.5 = 2.0 m
V4 = kQ/r = 9x109 x9.0x10-9 /2.0 = -40.5 V
(d) V' due to first charge, V' = kQ/r = 9x109 x-6.0x10-9/1.9 = -28.4 V
V" due to the 2nd charge, kQ/r = 9x109 x9.0x10-9 /2.42 = 33.46 V
Total potential V' + V" = -28.4 + 33.46 = 5.06 V
The distance in the last case is sqrt(1.9^2 +1.5^2) = 2.42 m
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