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-----problem-------

an infinite line of charge produces an electric field of 20N/C at a point that is 3 meters from the line. the electric field points toward the line of charge

what is would be the force (magnitude and direction) on a negative 6 microcoulomb charge that is located 6m from the line?

-----equations/constants----

i want to use E = force/q_test but the equation does not factor distance of 6m in

but if i use the F = kq_1q_2/r^2 equation i don't know the second charge

what should i do?

thanks

electric field line of charge E = lambda/2pi(epsilon_o)r where lambda = 3.34*10^-9C/m, epsilon_o = 8.85*10^-12, r = 6m

capacitance C = Q/V where q is charge, v is electric potential

electric field E = F/q_test = kq/r^2 where F is force, k = 9*10^9, r is distance = 6m

force F = kq_1q_2/r^2

-----attempt-----

if i use the E = F/q_test equation it does not allow me to factor in distance/radius r of 6m

but if i use the F =kq_1q_2/r^2, i don't know what the the second charge is

what equation should i use?

thanks

an infinite line of charge produces an electric field of 20N/C at a point that is 3 meters from the line. the electric field points toward the line of charge

what is would be the force (magnitude and direction) on a negative 6 microcoulomb charge that is located 6m from the line?

-----equations/constants----

i want to use E = force/q_test but the equation does not factor distance of 6m in

but if i use the F = kq_1q_2/r^2 equation i don't know the second charge

what should i do?

thanks

electric field line of charge E = lambda/2pi(epsilon_o)r where lambda = 3.34*10^-9C/m, epsilon_o = 8.85*10^-12, r = 6m

capacitance C = Q/V where q is charge, v is electric potential

electric field E = F/q_test = kq/r^2 where F is force, k = 9*10^9, r is distance = 6m

force F = kq_1q_2/r^2

-----attempt-----

if i use the E = F/q_test equation it does not allow me to factor in distance/radius r of 6m

but if i use the F =kq_1q_2/r^2, i don't know what the the second charge is

what equation should i use?

thanks

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