Answer to Question #165632 in Electric Circuits for EMMA HUGHENS

Question #165632

 A small positively-charged particle that has a charge of 1.20x10^-4 C came into the vicinity of a large positively-charged particle that created an electric force upon the test charge with an amount of 23.0x10 N. What is the Electric Field Strength exerted by the point charge then?


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Expert's answer
2021-02-22T10:19:56-0500

charge - q=1.2104Cq = 1.2*10^{-4}C Force - F=230NF = 230N

The Filed Strength exerted by the point is E=Fq=230N1.2104CE = \large\frac{F}{q}= \frac{230N}{1.2*10^{-4}C} =19.1(6)105NC= 19.1(6)*10^{5}\large\frac{N}{C}


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