26: PSS 26.1: The Electric Field of Multiple Point Charges

INTRO:
Three positively charged particles, with charges q₁=q, q₂=2q, and q₃=q (where q>0), are located at the corners of a square with sides of length d. The charge q₂ is located diagonally from the remaining (empty) corner.

Find the magnitude of the resultant electric field Enet in the empty corner of the square.

-----------------------------------------------------------------------------------------------------
PART A:
Below is an incomplete pictorial representation of the situation described in this problem. Complete the sketch by drawing the electric field due to each charge at point P. Make sure that all your vectors have the correct orientation.
The orientation of your vectors will be graded. The length of your vectors will not be graded.

SOLUTION:
Very obvious. The direction of the electric field is the same as the connecting line between the charge and point P

-----------------------------------------------------------------------------------------------------
PART B:
Now, consider the resultant electric field E_net at P. With reference to the coordinate system shown in the previous part, which component of E_net, if any, is zero in this problem?

• only the x component
• only the y component
• both the x and y components
• neither the x nor the y component

SOLUTION:
none of the components is zero, so the last option is correct
neither the x nor the y component

-----------------------------------------------------------------------------------------------------

PART C:

Determine the magnitude E_net of the net electric field at point P. Use K for the electrostatic constant.
Express your answer in terms of q, d, and K.

SOLUTION:

Using equation (26.1) E = kq/r², all components of all electric fields can be determined. 

the electric field due to q₃ at point P is just an x-component of a field...

the distance in consideration is just d, so
E₃ = E_3x =kq/d²

the electric field due to q₂ at point P is both an x-component and a y-component of a field...

since the setup is that of a square, the magnitude of the x-component is equal to the magnitude of the y-component. 
E_2x = E₂cos(45) = E₂cos(45) = E₂⋅1/√2 

E_2x = kq√2/(2d²)

the y-component is the same but in the y-direction.

E_netx = E_3x + E_2x 

= kq/d² + (√2/2)kq/d²
⇒E_netx =(1+√2/2)kq/d²

the y-magnitude of this net electric field is the same as the x-magnitude, but in the y-direction. 
total mag = √(x+y)

& if |x| = |y| ... then total mag = x⋅√2 = y⋅√2

→ E_net = √2⋅(1+√2/2)kq/d²

⇒ E_net =(1+√2)kq/d²

-----------------------------------------------------------------------------------------------------

PART D:

Intuitively, which of the following would happen to E_net if d became very large?

1. A) E_net should reduce to the field of a point charge of magnitude q.
2. B) E_net should reduce to the field of a point charge of magnitude 4q.
3. C) The larger d becomes, the smaller the magnitude of E_net will be.
4. D) The larger d becomes, the greater the magnitude of E_net will be.

Enter the letters of all the correct answers in alphabetical order. Do not use commas. For instance, if you think that A and D are correct, enter AD.

SOLUTION:

C) The larger d becomes, the smaller the magnitude of E_net will be

NOTE:

Looking at your answer derived in Part C, you should see that the magnitude Enet will decrease as d gets larger, just as you would expect. Your results do make sense! It is interesting to notice that the field will not reduce to that of a point charge (of magnitude 4q) unless the point P is moved farther away from the three charges while they themselves remain at their original positions.