Question

If Q = 2 coloumb and force on it is F = 100 newton, then the value of field intensity will be

• A. 100 N/C
• B. 50 N/C
• C. 200 N/C
• D. 10 N/C

Answer 1

The correct option is B 50 N/C

Answer is B.

Electric force on a charge q placed in a region o electric field intensity is E and it is given by F = qE.
In this case, F = 100 N and q = 2 C.
So, $E=\frac{F}{q}=\frac{100N}{2C}=50N/C$.
Hence, the value of field intensity will be 50 N/C.