Question

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:

1. a) What is the electric field at the location of the electron? Give magnitude and direction.

2. b) What is the electrostatic potential at the location of the electron?

3. 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.)

4. d) What is the work required, for an external agent to bring the electron to this location, from infinity?

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

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

  

and

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

Similarly ,

So,  

So,  

(a)

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

  

in vector notation

or,   

Where,

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

  

Electric field at O due to particle at B will be

  

  

  

  ​

Electric field at O due to particle at C will be

  

  

  

  ​

  

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

  

  

So, the magnitude of net electric field is

  

And the direction is

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

  

Where,

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

  

  

  

Electric potential at O due to particle at B will be

  

  

  

Electric potential at O due to particle at C will be

  

  

  

  

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

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

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

  

The magnitude of force is

  

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

   (potential at infinity is zero)

or,

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