1-5 please. A positive charge of magnitude 3.0 × 10-6 C is placed 5.0 em above t
ID: 1873713 • Letter: 1
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1-5 please.
A positive charge of magnitude 3.0 × 10-6 C is placed 5.0 em above the origin of a coordinate system, and a negative charge of the same magnitude is placed 5.0 cm below the origin both on the z-axis. What is the potential energy of a positive charge of magnitude 0.20 × 10-6C placed at the positi r. y. )(30 cm, 0 cm, 50 cm)? at (30 cm, 0 cm. 0 em)? . Consider two charges of 24 × 10-2C and-10x 10-2pC pectively, at opposite ends of a diameter of a circle of radius cm (Fig. 25-1). (a) What is the potential on a point of the le that is 30 cm from the positive charge? (b) How much rk is required to bring a charge of-0.2 ,C fron infinity to t point on the circle? -0.2 MC 10x10 O 24x10 A FIGURE 25 Problem 2 Charges of+12 pC and-20 C are placed along the y-axis at positions y = +5.0 cm and y =-9.0 cm respectively. (a) what is the potential at y = 0 and x 12.0 cm? (b) what is the poten- tial at y = 0, x-07(c) what is the electric field at that point? 3. 4t In a certain region of space, the electric potential due to a charge distribution varies only with x, changing according to V = a0 + air where ao12.7 V, al =-6.68 V/m, and x is in meters Find the electric field, magnitude, and direction in this region. A rod that is 20 cm long is given a uniformly distributed arge of 2 uC (Fig. 25-2). Calculate the potential at a point P which is a distance of 10 cm from the end of the rod, assuming that V = 0 at infinity.Explanation / Answer
1. q1 = 3 x 10^-6 C
q2 = -3 x 10^-6 C
(A) distance of q1 from P,
r1 =sqrt(30^2 + 0^2 + (50 - 5)^2) = 54.08 cm = 0.5408 m
from q2. r2 = sqrt(30^2 + 0^2 + (50 +5)^2) = 62.65cm = 0.6265 m
V = k q1 / r1 + k q2 / r2
(where k = 1/4 pi e0 = 9 x 10^9 )
V = 6829.5 Volt
PE = q V = (0.20 x 10^-6) (6829.5) = 1.37 x 10^-3 J .,....Ans
(b) P (30, 0 , 0)
this point is equidistant from both charges.
r1 = r2
hence V= 0
So PE = 0
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