Two point charges, q1 =-4.0 µC and q2 = 4.0 µC, are placed on the x axis at X1 = 4.7 m and x2-4.7 m, respectively (see figure below).=P92R(x2, 0)(X1, 0)(a) What are the electric potentials at the points P (0, 4.7 m) and R (9.4 m, 0)?Vp = 0VVREnter a number. differs from the correct answer by more than 10%. Double check your calculations. V(b) Find the work done in moving a 1.0-µC charge from P to R along a straight line joining the two points.(c) Is there any path along which the work done in moving the charge from P to R is less than the value from part (b)? Explain.Yes, because the work done depends only on the length of the path chosen.No, because the work done doesn't depend on the change in potential.Yes, because the change in potential will vary depending on the path chosen.No, because the work done depends only on the change in potential.Need Help?Read It

Given The charge at x1 is q1=-4 μC=-4×10-6 C. The charge at x2 is q2=4 μC=4×10-6 C. The distance…

Answered: Two point charges, q1 = -4.0 µC and q2…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Two point charges, q1 =
-4.0 µC and q2 = 4.0 µC, are placed on the x axis at X1 = 4.7 m and x2
-4.7 m, respectively (see figure below).
=
P
92
R
(x2, 0)
(X1, 0)
(a) What are the electric potentials at the points P (0, 4.7 m) and R (9.4 m, 0)?
Vp = 0
V
VR
Enter a number. differs from the correct answer by more than 10%. Double check your calculations. V
(b) Find the work done in moving a 1.0-µC charge from P to R along a straight line joining the two points.
(c) Is there any path along which the work done in moving the charge from P to R is less than the value from part (b)? Explain.
Yes, because the work done depends only on the length of the path chosen.
No, because the work done doesn't depend on the change in potential.
Yes, because the change in potential will vary depending on the path chosen.
No, because the work done depends only on the change in potential.
Need Help?
Read It

Expert Solution

Part (a)

Given

The charge at x₁ is $q_{1} = - 4 μC = - 4 \times 10^{- 6} C$ .
The charge at x₂ is $q_{2} = 4 μC = 4 \times 10^{- 6} C$ .
The distance between origin to x₁ is $x_{1} = 4.7 m$ .
The distance between origin to x₂ is $x_{2} = - 4.7 m$ . Here negative sign only shows that the point is on - x-axis.
The distance of the point P from the origin is $O P = 4.7 m$ .
The distance of the point R from the origin is $O R = 9.4 m$ .

The schematical diagram of the charges is shown as,

The distance between charges and point P is calculated as,

$\begin{array}{rcl} r_{1} & = & r_{2} = \sqrt{{(x_{1})}^{2} + {(O P)}^{2}} \\ = & \sqrt{{(4.7 m)}^{2} + {(4.7 m)}^{2}} \\ = & \sqrt{44.18 m^{2}} \\ = & 6.65 m \end{array}$

The potential at point P is calculated as,

$\begin{array}{rcl} V_{P} & = & \frac{k \times q_{1}}{r_{1}} + \frac{k \times q_{1}}{r_{1}} \\ = & \frac{k \times (- 4 \times 10^{- 6} C)}{6.65 m} + \frac{k \times (4 \times 10^{- 6} C)}{6.65 m} \\ = & 0 \end{array}$

Here, k is the coulomb constant whose value is $9 \times 10^{9} {Nm}^{2} / C^{2}$ .

The distance between the charge $q_{1}$ and the point R is calculated as,

$\begin{array}{rcl} d_{1} & = & O R - x_{1} \\ = & 9.4 m - 4.7 m \\ = & 4.7 m \end{array}$

The distance between the charge $q_{2}$ and the point R is calculated as,

$\begin{array}{rcl} d_{2} & = & O R + x_{2} \\ = & 9.4 m + 4.7 m \\ = & 14.1 m \end{array}$

The potential at point R is calculated as,

$\begin{array}{rcl} V_{R} & = & \frac{k \times q_{1}}{d_{1}} + \frac{k \times q_{1}}{d_{2}} \\ = & \frac{9 \times 10^{9} {Nm}^{2} / C^{2} \times (- 4 \times 10^{- 6} C)}{4.7 m} + \frac{9 \times 10^{9} {Nm}^{2} / C^{2} \times (4 \times 10^{- 6} C)}{14.1 m} \\ = & - 6000 Nm \end{array}$

Thus, the potential at point R is -6000 Nm.

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