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!
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