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A positive point charge Q1 = +Q is located at position (0,D)
and a negative point charge Q2 = -2Q is located at position
(L,D). What is the electric field at the point (x=L, y=0)?
Express your answer in unit vector notation using the
symbols given above and the constants k or 0.
y
L
Q1=+Q
D
Q2=-2Q
Q
Ek 2
r
D
x
What is the electric field at the point (x=L, y=0)?
E  E1  E 2   E1x  ˆi   E1y  E 2y  ˆj


E1x  E1 cos    k


y

Q

2 
2
2
L  D  

 

L
L2  D2


kQL


2
2 32
 L  D 

L
Q1=+Q
Q2=-2Q
D
E2
E1y 
D

E1

x
E2y
L
2
kQD
D
2kQ
 2
D

2 32
What is the electric field at the point (x=L, y=0)?

 

kQL
kQD
2kQ  ˆ
ˆ



E
i 
 2 j
3
2
3
2
  L2  D2     L2  D2 
D 

 

y
You could factor kQ out of the
expression on the right hand side,
but I don’t see that it simplifies
anything much, so let’s put a box
around our answer and call it done.
L
Q1=+Q
Q2=-2Q
D
D
E2

E1

x
A negative point charge -q is placed at (x=L, y=0). What is the
electric force on the point charge? Express your answer in unit
vector notation using the symbols above and the constants k
or 0.
y
L
Q1=+Q
Let’s be smart here…
Q2=-2Q
F  qE
D
D
-q

F   q  E from previous slide
x


 
 
kQL
kQD
2kQ  ˆ 

ˆ



F   q 
i 
 2 j
3
2
3
2
2
2
  L2  D2




D
L

D
   

 
 
Again, you could factor kQ out of the expression on the right hand side, or
multiply both terms by –q. How about instead let’s put a box around our
answer and call it done.
y
L
Q1=+Q
Q2=-2Q
D
D
-q
x