At a certain distance from a point charge, the magnitude ofthe electric field is

Question

At a certain distance from a point charge, the magnitude ofthe electric field is 630V/m and the electric potential is -2.40KV. (a) What is the distance to the charge? Round your answer to 3sig.figs( in m.) (b) What is the magnitude and sign of the charge? Round your answer to 3sig.figs.(in micro-Coulomb). Coulombconstant Ke=8.89*10^9 N.m^2/C^2 Please, all the posted problems are due tomorrow at 5:00pm C.Tby my professor and I do appreciate if someone could help with themall. Thank you so much for your help. At a certain distance from a point charge, the magnitude ofthe electric field is 630V/m and the electric potential is -2.40KV. (a) What is the distance to the charge? Round your answer to 3sig.figs( in m.) (b) What is the magnitude and sign of the charge? Round your answer to 3sig.figs.(in micro-Coulomb). Coulombconstant Ke=8.89*10^9 N.m^2/C^2 Please, all the posted problems are due tomorrow at 5:00pm C.Tby my professor and I do appreciate if someone could help with themall. Thank you so much for your help.

Explanation / Answer

the electric field is E = 630 V/m the electric potential is V = -2.40 kV = -2.40 * 10^3 V (a)the distance to the charge is V = E * d or d = (V/E) = (2.40 * 10^3/630) = 3.8 m (b)the magnitude of the charge is V = k * (q/d) or q = (V * d/k) V = -2.40 kV = -2.40 * 10^3 V,d = 3.8 m and k =(1/4o) = 9 * 10^9 Nm^2/C^2 or q = (-2.40 * 10^3 * 3.8/9 * 10^9) = -1013.3 * 10^-6 C =-1013.3 * 10^-6 C = -1013.3 C the magnitude of the charge is 1013.3 C and the sign ofthe charge is negative.

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