Two point charges are 10.0 cm apart and have charges of 2.0 muC and -2.0 muC, respectively. What is the magnitude of the electric field at the midpoint between the two charges?

1 Answer
Nov 9, 2017

"please look at the fallowing details."

Explanation:

  • Every electric charge has a electric field.
  • The charges in the electric field are affected each other.
  • As you move away from the electric charge, the intensity of the electric field decreases.
  • The electric field is a vector quantity.
  • The electric field of each charge is calculated to find the intensity of the electric field at a point.
  • The vectorial sum of the vectors are found.
  • The electric field of the positive charge is directed outward from the charge.

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  • the electric field of the negative charge is directed towards the charge.

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  • we can draw this pattern for your problem.

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"let the electric field of charge +2 "mu C" be " color(red)(E_1)" (red vector)"

d=5 cm=5.10^(-2)m
q_1=+2 mu C=+2*10^(-6)C

  • The electric field of a charge is calculated by formula :

E=k.(q)/(d^2)

color(red)(E_1=k*(2.10^(-6))/(5.10^(-2))=+k*0.4*10^(-4)N/C)

color(blue)(E_2=k*(-2.10^(-6))/(5.10^(-2))=-k*0.4*10^(-4)N/C)

  • The electric field of the charges at the point C is E_C

E_C=color(red)(E_1)+color(blue)(E_2)

E_C=k*0.4*10^(-4)+k*0.4*10^(-4)

E=2*k*0.4*10^(-4)=0.8*k*10^(-4)

k=9.10^9

E_C=0.8*9.10^9*10^(-4)

E_C=7.2*10^5 N/C