The charges q1= 18.5 micro Coloumb and q2= 24.0 micro Coloumb are fixed at points O and P, respectively on the base corners of the trapezoid shown below.a.) Compute the potential points M and Nb.) What is the work done to move a point charge of +2.5 micro Coloumb at constant speed from point M to point N. 4. The charges qı = - 18.5 µC and q2 = 24.0 µC are fixed at points O and P,respectively on the base corners of the trapezoid shown below. (a) Compute thepotential at points M and N. (b) What is the work done to move a point charge of+ 2.5 µC at constant speed from point M to point N.M0.223 m0.200 m60°

Given- q1 = -18.5 μC = -18.5 x 10-6 C , placed at O q2 =24.0 μC = 24.0 x 10-6 C , placed at P qo…

Answered: 4. The charges qı = - 18.5 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)...

See similar textbooks

Related questions

Concept explainers

Question

The charges q1= 18.5 micro Coloumb and q2= 24.0 micro Coloumb are fixed at points O and P, respectively on the base corners of the trapezoid shown below.

a.) Compute the potential points M and N

b.) What is the work done to move a point charge of +2.5 micro Coloumb at constant speed from point M to point N.

4. The charges qı = - 18.5 µC and q2 = 24.0 µC are fixed at points O and P,
respectively on the base corners of the trapezoid shown below. (a) Compute the
potential at points M and N. (b) What is the work done to move a point charge of
+ 2.5 µC at constant speed from point M to point N.
M
0.223 m
0.200 m
60°

Expert Solution

Step 1- Analyzing problem

Given-

q₁ = -18.5 $μ$ C = -18.5 x 10^-6 C , placed at O

q₂ =24.0 $μ$ C = 24.0 x 10^-6 C , placed at P

q_o =2.5 $μ$ C = 2.5 x 10^-6 C

Step 2-(a) Determination of potential at M

fig-1

From fig-1,

$O Q = O M \cos 60 ° = 0.200 \cos 60 ° = 0.1 m M Q = O M \sin 60 ° = 0.200 \sin 60 ° = 0.173 m$

$M Q = N P = 0.173 m a n d M N = Q P = 0.223 m$

$O P = O Q + Q P = (0.1 + 0.223) m = 0.322 m$

$f o r r t . a n g l e △ M Q P, M P = \sqrt{M Q^{2} + Q P^{2}} M P = \sqrt{0 . 173^{2} + 0 . 223^{2}} m M P = 0.282 m$

$f o r r t . a n g l e d △ N P O, O N = \sqrt{N P^{2} + O P^{2}} O N = \sqrt{0 . 173^{2} + 0 . 323^{2}} m O N = 0.366 m$

Let the potential at point M be V_M , then

$V_{M} = \frac{k q_{1}}{O M} + \frac{k q_{2}}{M P} V_{M} = (\frac{9 \times 10^{9} \times - 18.5 \times 10^{- 6}}{0.2}) + (\frac{9 \times 10^{9} \times 24 \times 10^{- 6}}{0.282}) V_{M} = - 66542.553 V$

Hence, potential at point M is -66542.553 V

Step by step

Solved in 4 steps with 1 images

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.

Similar questions