Electric field at a distance of 20cm from a 4µC charge is:
Correct answer: C. 9 x 10^5N/C
- A. 3 x 10^5N/C
- B. 4.5 x 10^5N/C
- C. 9 x 10^5N/C
- D. 3 x 10^3N/C
Explanation
The electric field E due to a point charge q at a distance r from the charge is given by Coulomb's law: E=kq /r2where:k is Coulomb's constant = 8.99×109Nm2/C2q is the charger is the distance from the charge.Substituting the given values (q = 4µC = 4x10−6C) & (r = 20cm = 0.2m) into the equation, we get:E=(8.99×109Nm2/C2)×(4×10−6C )/(0.2m)2 =(8.99×109Nm2/C2)×(4×10−6C )/(0.04m2 or 4×10-2m2)=8.99×109×10-6×102 N/C =8.99×105N/C≈9x105 N / C
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About Electric Field Intensity
Electric field intensity is the force acting per unit positive test charge, with direction set by the force on that charge. Work includes fields of point charges, vector superposition, field lines and the relation between field and electric potential. Field intensity must be distinguished from electric force and potential.
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