Question

o Show enough work to guide your reader through the steps (but not every gory detail of arithmetic or algebra is needed). o Draw the diagrams necessary to apply the principles (in this case vector diagrams-whatever vectors are calculated must be drawn and supported with triangles). Hand write your solution, or use Mathcad. Do not use a word processing application for math (too time consuming). If using Mathcad, hand-draw the figures. o Include units on unit-bearing quantities, even within calculations o Attach this page as a cover sheet PROBLEM Three charged particles, labeled 1,2, and 3, are fixed in place at the following coordinates: particle 1 (0 m,0.5 m), particle 2 (0.5 m, -0.5 m), and particle 3 (-0.5 m, -0.5 m). The charges for the particles are as follows: particle 1: -2 nC; particle 2:-1 nC; particle 3: 3 nC. An electron, free to move, is introduced at coordinates (0.2 m,-0.2 m). Address the following questions a) What is the electric field at the location of the electron? Give magnitude and direction b) What is the electrostatic potential at the location of the electron? c What is the electrostatic force on the electron? Give magnitude and direction. (Hint: apply one or more of your results from above, rather than starting this calculation from scratch.) What is the work required, for an external agent to bring the electron to this location, from infinity? d)

**Show all work and equations used, provide diagrams if needed

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Answer #1

Given,

Charge on particle 1 (q_{1}) is -2 n C i.e. -2 x 10-9 C

charge on particle 2 (q_{2}) is -1 x 10-9 C

charge on particle 3 (q_{3}) is 3 x 10-9 C

1 (0,0.5) e (0.2 ,-0.2) 2 (0.5 ,-0.5)

From this diagram we can calculate the vectors overrightarrow{AO} ,  во , and  CO . Here, A , B , C , and O is the position of particle 1 , 2 , 3 , and the electron respectively.

AÔ = (position vector of point O)-( position vector of point A)

  =( 0.2 , hat{i} - 0.2 , hat{j}) -(0 , hat{i} + 0.5 , hat{j})

0.2i - 0.7

and R = V(0.2)2 + (-0.7)2 = 0.728 m

(Position vector of a point is defined as the vector starting from the origin and ending to the point. )

Similarly ,

B0-(0.22-0.2 j) _ (0.52ー0.5 j)ーー0.3 i + 0.3j

So,  BỎ| = V/(-0.3)2 (0.3)-= 0.424 m

^ CO = (0.2 i-0.2 j) _ (-0.5 -0.5 j)-0.7 ị t 0.3 j

So,  CO- V(0.7)+ (0.3)2 -0.7615 m

(a)

Electric field due to a charged particle having charge ' q ' at a distance ' r ' is given by

  E = rac{1}{4 pi epsilon _{0}} , rac{q}{r^{2}}

in vector notation

vec{E} = rac{1}{4 pi epsilon _{0}} , rac{q}{r^{2}} , rac{vec{r}}{left | vec{r} ight |}

or,   vec{E} = rac{1}{4 pi epsilon _{0}} , rac{q}{r^{3}} ,vec{r}

Where,

4TTEQ

Using this equation we can calculate the electric field due to all the three particles at O.

So, Electric field at O due to particle at A will be

  1 41

  2.0 × 10-9 07283一一x(0.2 i-07j) 9 × 109

ーー46.653 × (0.2 i-0.7j)

9.331 + 32.657

Electric field at O due to particle at B will be

  1 92

  = 9 × 109 × × (-0.3i +0.3j) 0.4243

  -1 18.07x (-0.3 ị t 0.3)

  35.421-35.42 )

Electric field at O due to particle at C will be

  1 93

  109 × 3.0 × 10-. x (0.7i+0.3j) 0.76153

  61.144 x (0.70.3j)

  42.82 + 18.34 j

  

So, net electric field at point O due to all the three particles is

vec{E} = vec{E}_{AO} + vec{E}_{BO} + vec{E}_{CO}

  ( -933 i + 32.6575) + (35.42 i-35.425) + ( 42.8 i + 18.34 j )

  68.891-15.577

So, the magnitude of net electric field is

  E- E (68.89)(15.5772V4988.47 70.63V/m

And the direction heta is

-tan-ı 15.577 15.577 68.8912.74

Hence, the net electric field is making an angle of 12.740 (or nearly 130) with the x - axis.

(b)  

Electric potential due to a charged particle having charge ' q ' at a distance ' r ' is given by

  V = rac{1}{4 pi epsilon _{0}} , rac{q}{r}

Where,

4TTEQ

Using this equation we can calculate the electric potential due to all the three particles at O.

So, Electric potential at O due to particle at A will be

  1 91 VÁo 4πέρ |.AO

  2.0 x 10-9 0,728 9 × 109

  = 9 imes (-2.747) , , V

Electric potential at O due to particle at B will be

  V_{BO} = rac{1}{4 pi epsilon _{0}} , rac{q_{2}}{left | overrightarrow{BO} ight | }

  = 9 imes 10^{9} imes , rac{-1.0 imes 10^{-9}}{0.424}

  = 9 imes (-2.358) , , V

Electric potential at O due to particle at C will be

  V_{CO} = rac{1}{4 pi epsilon _{0}} , rac{q_{3}}{left | overrightarrow{CO} ight | }

  .. 3.0 × 10-9 =9×109 ×-07615

  = 9 imes (3.94) , , V

  

So, net electric potential at point O due to all the three particles is

V = V_{AO} + V_{BO} + V_{CO}

9 × (-2.747-2.358+ 3.94)--10.485 1

(c) The force on a charge particle (having charge ' q ') in electrostatic field ' E ' is given by

vec{F} = q vec {E}

So, the force on electron ( charge on electron = -1.602 x 10-19 C) at the point O will be

  F-(-1.602 × 10-19) × (68.892-15.577

-110.36 × 10-19 i-24.954 × 10-19j

The magnitude of force is

  19 F-İFİ-V(-110.36 10-19)2+ (24.945x 10-19 )2-1 13.15x 10- Ņ

And the direction of the force will be in the opposite direction of electric field because electron has negative charge.

(d) The work done in bringing the electron from infinity to the point O will be

  inf inity (potential at infinity is zero)

or, Wーー1.602 × 10-19 × ((-10.485)-0) = 16.8 × 10-19 J

For any doubt please comment and please give an up vote. Thank you.

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