Three point charges are arranged in a horizontal line as shown below. Find the e
ID: 1587261 • Letter: T
Question
Three point charges are arranged in a horizontal line as shown below. Find the electric forces (in units of kQ2/R2) on the charges given the following: Q1 = 100 Q, Q2 = -5 Q, Q3 = 36 Q, r1 = 5 R, and r2 = 1 R. Remember that a positive force points to the right and a negative force points to the left.
What is the net force on charge Q1?
_________kQ2/R2
What is the net force on charge Q2?
______kQ2/R2
What is the net force on charge Q3?
______kQ2/R2
What is the sum of the forces on all three charges?
______ kQ2/R2
Explanation / Answer
Unfortunately their figure doesn't look, but with data from your writing I'm going to build the graph
+Q3 +Q1 - Q2
. r1 r2
Data
Q1 = 100 Q
Q2 = - 5Q
Q3 = 36 Q
. r1 = 5R
. r2 = 1 R
Part A net force about Q1
--------- > F12
-------- > F13
+Q3 + Q1 - Q2
. r1 r2
We calculate the power using the sum with Newton's law and coulomb's law
Fn= F12 + F13
Fn = k q1 q2 / r22 + k q1 q3 / r12
We substitute the given values
Fn = k q2 q1/ (R)2 + kq1 q3/ (5R)2) Fn = k ( 5Q 100Q/R2 + 100Q 36Q/25R2
Fn = k (500QQ/R2 + 144QQ/ R2 Fn = kQ2 /R2 ( 244 )
Part B
F32 < --------
F12 < ----------
+Q3 + Q1 - Q2
. r1 r2
Fn= - F32 – F12
Fn = - k q3 q2/(r1 +r2)2 – k q1 q2/r22 Fn = - k ( 36Q 5Q /(5R+R)2 + 100Q 5Q/ R2)
Fn = - k Q2/R2 ( 180QQ/62 + 500Q2 ) Fn = - k Q2 /R2 ( 505)
Fn = - k Q2/R2 (-505)
Part C
F32 -------- >
F31 <-------
+Q3 + Q1 - Q2
. r1 r2
Fn= F32 – F31
Fn = k q3q2/(r1+r2)2 - k q3 q1/r12 Fn = k ( 5Q 36Q/(5R+R)2 – 36Q 100Q/(5R)2
Fn = K (5QQ/R2 - 3600/25 Q/R2 Fn = k Q2/R2 (5 - 144)
Fn = k Q2/R2 ( - 139)
Part d
Join all the previous results
Fn =k Q2/R2 (244 -505 - 139)
Fn = k Q2 /R2 ( -400 )
I hope that this is the case any other setting should only change the signs
Another possible configuration is
+Q1 +Q3 -Q2
. r1 r2
Part A (r1+r2) = 5R+R = 6R
------ > F12
------ > F13
+Q1 +Q3 -Q2
. r1 r2
Fn = F12+ F13 Fn = k q1q2/(r1+r2)2 + k q1 q3/r1 Fn = k (q1q2/(6R)2 + q1q3/q2 R2)
Fn = k /R2 (100Q 5Q/36+100Q 36Q ) Fn= k Q/R2 3613.9
Part B
<------ F23
<------ F21
+Q1 +Q3 -Q2
. r1 r2
Fn= - F23- F21= - ( F23+F21) Fn = - k (q2q3/r22 + q1q2/(r1+r2)2)
Fn = - k ( 5Q 36Q/R2 +100Q5Q/36R2) Fn = k Q/R2 (- 193Q)
Part C
<--- --- F13
------ > F32
+Q1 +Q3 -Q2
. r1 r2
Fn = F32 – F13 Fn = k ( q3q2/r22 – q1 q3/(r1+r2) Fn = k ( 36Q5Q/R2 – 100Q 36Q/ 36R2)
Fn = k Q2 /R2 ( 180– 100) Fn = k Q2 /R2 (80)
I think that these are the two most probable configurations
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