The electric field intensity at a point 20cm away from a charge of 2 x 10^-5C is:
Correct answer: A. 4.5 x 10^6 N/C
- A. 4.5 x 10^6 N/C
- B. 3.5 x 10^5 N/C
- C. 3.5× 10₀ N/C
- D. 4.5 x 10^5 N/C
Explanation
E = F/q Where F is the force and q is the chargeSo,E = kq2/qr2 Where K is the value of 1/(4πεo)E = kq/r2E = (9x109 x 2x10-5) / (0.2)2E = 4.5x105 NC-1
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