# What is the force exerted on q1, q2, q3 by the other two chargesI can’t solve b and c. I’ve tried answers, but they keep appearing wrong. Please help me get the right answers.Please use the...

What is the force exerted on q1, q2, q3 by the other two charges

Please use the Coulomb constant given, convert the cm to m, and be aware of positive/negative signs. Some of the answers are negative numbered answers, and I suspect I may be getting wrong answers because of the signs.
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Three point charges, q1 = +6.3 µC, q2 = +2.6 µC, and q3 = −3.9 µC, lie along the x-axis at x = 0 cm, x = 2.0 cm, and x = 5.5 cm, respectively. The Coulomb constant is 8.99 ×10^9 N • m^2/C^2.

b) What is the force exerted on q2 by the other two charges? (To the right is positive.) Answer in units of N.
c) What is the force exerted on q3 by the other two charges? (To the right is positive.) Answer in units of N.

neela | High School Teacher | (Level 3) Valedictorian

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We know that the force F between the two charges q1and q2 separated by a distance d is given by:

F = [1/(4pi*e0)]*(q1*q2/d^2). Please do not get confused with the  notation, as we use this as a formula throughout to calculate the force between the two charges separated by a distance.

a) Let the postion of 3 charges charges placed be at O(0,0), A(2,0), B(5.5,0). Then the force exerted on q2 by the other two charges is given by:

F2 = q2*q1/[(4pi*e0*)*((2-0)cm)^2] + q2q3/[4pie0*((5.5-2)cm)^2 ]

= (q2*q1*2.5^2 right (repulsion on q2 by q1) +q2*q3*2^2 right (attraction on q2 by q3))/[4pi*e0*(2*2.5)^2]

={(6.3*2.6*6.25 right +2.6*(+3.9)*2^2 ) right}/ (4pie0*5^2*cm^2}

={(6.3*2.6*6.25 right +2.6*(3.9)*2^2 ) right}/ (4pie0*5^2*100^(-2)m^2}

= 142.935 *10^-12*8.99*10^9*100^2 /25

=142.935*8.99*10

= 513.99426N

b)

The force exerted on q3 by the other two charges. Since q3 is -ve and the q1 and q2 are +ve, both q1 and q2 exert attractional force on Q3 is therefore, it ltowards left.

= q3q1/[4pie0*(5*5-0)^2]+q3*q2/(4pie0*(5.5-2)^2

= [1/(4pe0)]{(3.9)*(6.3)/((5*5-0) cm)^2 towardsO+(3.9)*(2.6)/((5.5-2)cm)^2 to wards A] *mu^2 towards left

=8.99*10^9{3.9*6.3*3.5^2+3.9*2.6*5.5^2}(muC^2)^2*100^2/(5.5^2*3.5^2) left

=8.99*10^9*{607.7175}*10^(-12)*10^4/306.25 N left

=178.3960922 N left

=-178.3960922N