a.)
Electric field is given by
E = k*q/r^2
r = distance between charges
q = magnitude of charge
k = 9*10^9
and direction of E will be towards -ve charge and away from +ve charge
given, q1 = 5.70C = 5.70*10^-6 C
q2 = 1.40C = 1.40*10^-6 C
q3 = -2.00C = -2.00*10^-6
C
r1 =3.00 cm = 3.00*10^-2 m
r3 = 2.00 cm =2.00*10^-2 m
r' = 1.00 cm = 1.00*10^-2 cm
so, electric field due to charge q1 is towards +x-axis and is given by ,
E1 = k*q1/(r1 - r') ^2= (9*10^9)*(5.70*10^-6)/(3*10^-2 - 1*10^-2)^2
E1 = 1.28*10^8 N/C
electric field due to charge q2 is towards -x axis and is given by,
E2 = k*q2/r'^2 = (9*10^9)*(1.40*10^-6)/(1.00*10^-2)^2
E2 = 1.26*10^8 N/C
electric field due to charge q3 towards +x-axis and given by,
E3 = k*q3/(r3 + r')^2 = (9*10^9)*(2*10^-6)/(2*10^-2 + 1*10^-2)^2
E3 = 0.2*10^8 N/C
So, net electric field on r' distance to the left of center marble is,
E = E1 + E3 2 E2
E = (1.28+0.2-1.26)*10^8
E = 2.2*10^7 N/C
b.) Since electric force on charge is given by,
F = q*E
here, E = electric field at that point = 2.2*10^7 N/C
q = 3.90 C = 3.90*10^-6
So, F = (3.90*10^-6)*(2.2*10^7)
F = 85.8 N
Please upvote.
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