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What is the magnitude of the force on a -2.8 µC charge placed at this point? Answer in...
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
High School Teacher
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
Posted by neela on February 20, 2010 at 2:24 PM (Answer #1)
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