Homework Help

What is the magnitude of the force on a -2.8 µC charge placed at this point? Answer in...

user profile pic

azura-huskie | Student, Undergraduate | eNotes Newbie

Posted February 20, 2010 at 5:27 AM via web

dislike 0 like

What is the magnitude of the force on a -2.8 µC charge placed at this point? Answer in units of N.

 

I solved the first part, but I can't solve the 2nd part of this question. Since I can't put a picture on here, I'll describe the picture:

q1, q2, and q3 on horizontal line, respectively.

q1 = +7.2 µC

q2 = +3.2 µC

q3 = -2.8 µC

Distance b/w q1 and q2 = 4.7 cm

Distance b/w q2 and q3 = 2.2 cm

--

SOLVED: What is the magnitude of the electric field strength at a point 2.2 cm to the left of the middle charge? ANSWER: 57113280.55 N/C

--

b) What is the magnitude of the force on a -2.8 µC charge placed at this point? Answer in units of N.

--

*The value of the Coulomb constant is  8.98755 ×10^9 N • m^2/C^2.

*Please change cm to m.

*PLEASE DO NOT ROUND NUMBERS UNTIL THE VERY END!! If there's a lot of numbers after decimal, please round it to the 6th or 7th place.

Please help me get the answer and thank you!

1 Answer | Add Yours

user profile pic

neela | High School Teacher | Valedictorian

Posted February 20, 2010 at 2:24 PM (Answer #1)

dislike 0 like

The force of attraction between the two charges Q1 and Q2 separated by a distance r  is given by:

F = (1/4pe0) Q1*Q2/r^2 N = 8.98755*10^9* Q1*Q2/r^2.

The force is of attraction if both chages are of opposite type or unlike charges and is of  repulsion if the charges are like ones.

In the given case:

The force  q1  exerted on q3 :

q1 = 7.8*10^-(6) .=C  and   q2 = 2.8*10^-6 C, r = (4.7+2.2)cm = 6.9 cm =6.9/100  = 0.069 metr.

The force q1 on q3 is given by:

F1 = 8.98755(7.8*10^-(6)*2.8*10^-6))/[(4.7+2.22)/100]

=38.0569225 N is attrction by q1 on q3 as they are unlike charges and is therefore towards left.

Similarly the force exeterted by q2 on q3:

q2 = 3.2*10^(-6) C and q3  = 2.8*10^(-6) and distance =2.2cm =2.2/100 meter

F2 = 8.98755*(3.2*10^(-6)*2.8*10^(-6)/(2.2/100) N

=166.38100909N attraction towards left as the charges attr unlike.

F1 and F2 are both  along X axis towards left and in the same direction. Therefore,

The exertion of force of charges q1 and q2 on q3 =sqrt [(F1)^2+(F2)^2+2F1*F2 cosine (angle between F1and F2)} = sqrt[( F1)^2+(F2)^2+2f1F2 cos0] = F1+F2 = 38.0569225 N + 166.38100909N

= 204.4380134 N towards left.

= - 204.4380134 N

 

 

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes