Question Solved1 Answer What magnitude point charges creates a 10000 N/C electric field at a distance of 0.5 m? a. 877.2 nc Ob. 287.7 nc Oc. 277.8 nc O d. 872.7 nc A force of 100 N exists between two identical point charges separated by a distance of 40 cm. What is the magnitude of the two point charges? O a. 100 LC b.42 pc O c. 40 nc Od. 50 mc Two charges 91 = 10 C and 92 = 5 pc are located along X and Y coordinate system. Charge q1 is along X-axis at a distance of 2 cm and q2 is along Y- axis at a distance of 4 cm. What is the magnitude of the total electric field at the origin of the coordinate system? a. 2.81 x 107 NC Ob.2.25 x 107 N/C Oc 2x 108 NC d. 2.26 x 108 N/C

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Transcribed Image Text: What magnitude point charges creates a 10000 N/C electric field at a distance of 0.5 m? a. 877.2 nc Ob. 287.7 nc Oc. 277.8 nc O d. 872.7 nc A force of 100 N exists between two identical point charges separated by a distance of 40 cm. What is the magnitude of the two point charges? O a. 100 LC b.42 pc O c. 40 nc Od. 50 mc Two charges 91 = 10 C and 92 = 5 pc are located along X and Y coordinate system. Charge q1 is along X-axis at a distance of 2 cm and q2 is along Y- axis at a distance of 4 cm. What is the magnitude of the total electric field at the origin of the coordinate system? a. 2.81 x 107 NC Ob.2.25 x 107 N/C Oc 2x 108 NC d. 2.26 x 108 N/C
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Transcribed Image Text: What magnitude point charges creates a 10000 N/C electric field at a distance of 0.5 m? a. 877.2 nc Ob. 287.7 nc Oc. 277.8 nc O d. 872.7 nc A force of 100 N exists between two identical point charges separated by a distance of 40 cm. What is the magnitude of the two point charges? O a. 100 LC b.42 pc O c. 40 nc Od. 50 mc Two charges 91 = 10 C and 92 = 5 pc are located along X and Y coordinate system. Charge q1 is along X-axis at a distance of 2 cm and q2 is along Y- axis at a distance of 4 cm. What is the magnitude of the total electric field at the origin of the coordinate system? a. 2.81 x 107 NC Ob.2.25 x 107 N/C Oc 2x 108 NC d. 2.26 x 108 N/C
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Sol: Given  A) Electric field E = 10000 N/C Distance r = 0.5 m We know that due to point charge electric field at a distance r is  E = kQ /r²........................(1) Where  K = 9*10^9 Nm²/C² So, from equation(1) Q = Er²/k     = (10000)(0.5)²/(9*10^9)     = 2.77777*10^-7      = 277.8*10^-9 C     ≈ 277.8 nC B) Force F = 100 N Distance r = 0.4 m We have force F = kQ1Q2/r² But here given identical charges  so Q1=Q2 = Q F = kQ²/r² Q&#1 ... See the full answer