Three point charges each of 4micro coulomb are placed at the three corners of a square of side 20cm.Find the magnitude of the force on each.

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Three point charges each of 4micro coulomb are placed at the three corners of a square of side 20cm.Find the magnitude of the force on each.

Physics
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It is much long question. :/ Well I try to do.
thank you shayaan plz help me to do it.
You have to wait for this. Cause Its my pray time and when I will come back to home then there will be no light at home. According to PST(Pakistan standard time) at 3pm light will back then I will post it. InshALLAH. Tc.

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Let's find the force on charge 1 we know Force between two points charges = K*C1*C2/r^2 K=9*10^9 Nm^2/C^2 Force due to 2 F12= 9*10^9*4*4*10^-6/(20*10^-2)^2 =3.6*10^6 N Force due to charge 2 on 1 is in upward direction similarly force due to charge 3 on charge 2 will be F13=3.6*10^6 N but in the left direction|dw:1327309283490:dw| the resultant force will be \[F1=\sqrt{{F12}^2+{F13}^2}\] F1= 5.019*10^6 in the direction see the diagram|dw:1327309521627:dw|
Force on Charge 2 Force due to Charge 1 F21=3.6*10^6 N this force will be in the downward direction Length of the diagonal= root(20^2+20^2)=28.28 cm Force due to the charge 3 F23= k*4*4*10^-6/(28.28*10^-2)^2 F23=1.8*10^-6 N in the direction shown in the diagram |dw:1327309831876:dw| F23 will have two components F23 cos 45 degree in left and F23 sin 45 in downward direction F23 cos 45=F23 sin 45=1.273*10^6 newtons so resultant force on charge 2 will be \[ F2= \sqrt{{(F23*sin 45+F21)}^2+{(F23cos 45)}^2}\] F2= 5.036*10^6 N |dw:1327310650713:dw|
Force on charge 3 will be of the same magnitude as on charge 2 F3=F2=5.036*10^6 it's direction is shown in the diagram |dw:1327310913317:dw|
|dw:1327313961419:dw| Magnitude of all three charges are +4μC. Due to +ve sign all charges repel each other. First Consider Charge 1: Hence two forces acts on it 1)F3 at 45° 2) F2 along +ve x-axis, On X-axis: F1x=F2x+F3xcos45 F1x=kq2q1/r² + kq3q1/r² cos45 By substituting values you will get your final answer as, F1x=6.145N On Y-axis: F1y=F3ysin45 F1y=kq3q1/r² sin45 By substituting values you will get your final answer as, F1y=2.545N F1=sqrt(F1x²+F1y²) F1=6.65N Second Consider Charge 2: Hence, two forces act on it 1)F1 2)F3 On X-axis: F2x=kq2q1/r² By substituting values you will get your final answer as, F2x=3.6N On Y-axis: F2y=kq2q3/r² By substituting values you will get your final answer as, F2y=3.6N F2=sqrt(F2x²+F2y²) F2=5.09N Third Consider Charge 3: Hence, two forces act on it 1)F2 along y-axis 2)F1 at 45° On X-axis: F3x=kq3q1/r² sin45 F3x=3.6N On Y-axis: F3y=F1ycos45+F2 F3y=kq2q3/r² cos45 + kq2q3/r² F3y=6.14N F3=sqrt(F3x²+F3y²) F3=7.11N
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sorry i took them as 4 mC , drop 10^6 from each force, that'll give you your answer

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