Electric field intensity

What is the intensity of the electric field at a point that lies in the middle between two point charges Q1 = 5.10-5 C and Q2 = 7.10-5 C, which is separated by r = 20 cm? The charges are placed in a vacuum.

Final Answer:

E =  -1797510.5235 V/m

Step-by-step explanation:

Q1=5 1050.0001 C Q2=7 1050.0001 C ε=8.854187 1012=8.85421012 F/m  r=20 cm m=20:100  m=0.2 m r2=r/2=0.2/2=101=0.1 m  E1=4π ε1 r2Q1=4 3.1416 8.854210121 0.10.00014493776.3087 V/m E2=4π ε1 r2Q2=4 3.1416 8.854210121 0.10.00016291286.8321 V/m  E=E1E2=4493776.30876291286.8321=1797510.5235 V/m1.798106 V/m



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